In this answer I will try to resolve your apparent contradiction first by showing arithmetically why only 1 outcome is true and in the next part try to convince your intuition why this is the case.
Arithmetic proof
As alluded to already in other responses, there is no contradiction between the Van't Hoff equation and the equation relating $K$ and $\Delta G$. Given:
\begin{align}
K &= \exp\left({-\Delta G^° \over RT} \right). \label{5-7}\tag 1\\
\end{align}
Assuming that $\Delta H^\circ$ and $\Delta S^\circ$ are invariant with temperature, we can differentiate the expression with respect to T to obtain:
\begin{align}
\frac{\mathrm{d}K}{\mathrm{d}T} &=
\frac{\Delta H^\circ\exp\left(
\frac{\Delta S^°}{R}-\frac{\Delta H^°}{RT}
\right)}{RT^2}
\label{four}\tag 2\\
\end{align}
Since $RT^2 >0$ and $\exp(\frac{\Delta S^°}{R}-\frac{\Delta H^°}{RT}) >0$, the sign of $\frac{\mathrm{d}K}{\mathrm{d}T}$ is only dependent on $\Delta H^\circ$. For exothermic reactions, increasing T will lead to decreasing $K$. Similarly, for endothermic reactions, increasing T will lead to increasing $K$. In other words, Le Chatelier's principle and the Van't Hoff equation are correct.
Intuition
This result might contradict our intuition. For a reaction with $\Delta H^\circ < 0$ and $\Delta S^\circ > 0$, we would expect, given the maths, that $K$ decreases with increasing temperature. However, as you mentioned, the negative $\Delta S^\circ$ term would mean $\Delta G^\circ$ decreases with increasing temperature. This gives rise to an apparent contradiction: at increasing temperatures, $G^\circ$ of products decreases relative to reactants, meaning (at standard state) products are more stable than reactants, yet the position of equilibrium is shifting toward the reactants.
The key to this conundrum lies in the $RT$ term and more specifically what temperature actually means. Notice that in equation (1), changing $T$ does not only change $\Delta G^\circ$ but also $RT$.
Let us consider just enthalpy first
For the sake of discussion let us consider the following: a closed box filled with inert gas with some molecules of A(g) which can isomerise to B(g):
$$\ce{A(g) <=> B(g)}$$
This system has constant total energy $E_{total}$. We consider each molecule of A or B as a smaller subsystem with energy $e_A$ and $e_B$ respectively. The energy of the rest of the system with respect to that one molecule is therefore $E_{total} - e$.
Consider that the isomerisation reaction is exothermic such that when $\ce{A(g) -> B(g)}$, heat is released such that $e$ decreases and $E_{total} - e$ increases. Consider also, for now, that the reaction involves no change in entropy ($\Delta S^\circ = 0$).
Hopefully it is intuitive that at equilibrium at some temperature $T_1$, more molecules of B than A exist, therefore $K > 1$. You can think of it this way: the B isomer contains stronger intramolecular bonds than A is therefore more stable than A. This is also why the reaction is exothermic.
Here is another more useful way to appreciate why $K > 1$. We consider the system as a microcanonical ensemble, such that every microstate occurs with equal probability. Let us rewrite the equation for $K$. Since $\Delta S^\circ = 0$:
\begin{align}
K &= \exp\left({-\Delta H^° \over RT} \right)
\end{align}
\begin{align}
K &= \frac{\exp\left({-H^°_B \over RT} \right)}
{\exp\left({-H^°_A \over RT} \right)}
\end{align}
\begin{align}
K \propto \frac{\exp\left({-H^°_B \over k_BT} \right)}
{\exp\left({-H^°_A \over k_BT} \right)} \tag{3}
\end{align}
Now, by considering the equation for Boltzmann's distribution,
\begin{align}
K \propto \frac{\text{Probability(molecule exists as B isomer)}}
{\text{Probability(molecule exists as A isomer)}}
\end{align}
\begin{align}
K \propto \frac{\Omega(E_{total} - e_B)}
{\Omega(E_{total} - e_A)}
\end{align}
Where $\Omega(E)$ is the number of microstates accessible to the macrostate with energy $E$.
From the re-expression we can appreciate that the equilibrium constant depends on the probability that the molecule exists as either A or B, which depends on the number of microstates available. Here is the key: because $e_A > e_B$, then $(E_{total} - e_A) < (E_{total} - e_B)$, such that $\Omega(E_{total} - e_A) < \Omega(E_{total} - e_B)$. Therefore, by virtue of the fact that the reaction is exothermic, $K>1$.
Now, there are 2 ways to appreciate that increasing the temperature of the system causes $K$ to decrease.
The simplest way is to appreciate that from equation (3), increasing T will cause both the numerator and denominator to approach 1, meaning $K$ decreases from $K > 0$ to $K=1$. But how does this work intuitively?
Let us consider the thermodynamic definition of temperature. For our system,
$$T = \frac{dE}{dS}$$
It is easier to consider the definition of temperature as:
$$\frac{1}{T} = \frac{dS}{dE}$$
Therefore, higher temperatures mean that the same increase in energy results in a smaller increase in entropy (in other words, a smaller increase in the number of possible microstates). In other words, a small decrease in energy results in lower entropic cost. Applying this to our system, we see that at higher and higher temperatures, $E_{total}$ increases, such that $(E_{total} - e_B) \approx (E_{total} - e_A) \approx E_{total}$. In other words, as temperature increases, the relative entropic cost of the molecule existing as A or B becomes negligible, to the extent that
$$\text{Probability(molecule exists as B isomer)} \approx \text{Probability(molecule exists as A isomer)}$$
such that $K=1$.
The above is generalizable to endothermic reactions. In the case where $\Delta H^\circ > 0$, at equilibrium at some temperature T more A than B molecules will be present such that $K<1$. As temperature increases, relative entropic cost of converting between A and B decreases such that $K=1$
Now let us consider entropy as well
Without considering entropy, the apparent contradiction does not exist. Therefore, let us consider that the reaction
$$\ce{A(g) -> B(g)}$$
Involves a positive entropy change. For the sake of simplicity, consider that this entropy change does not affect the entropy of the rest of the system. For example, the A isomer forms a closed, rigid ring while the B isomer forms a long linear molecule. (It is not always the case that $\Delta S^\circ$ of a reaction does not affect entropy of the system. Consider kosmotropic ions that reorganise solvent molecules into an ordered clathrate cage.)
How does this affect what have concluded so far for enthalpy? The answer: It doesn't change anything. As we increase temperature, the relative energies of A and B are still defined by the $\Delta H^\circ$. Therefore, increasing temperature means the relative entropic cost of converting between A and B diminishes, such that $K$ decreases.
The difference here is that $\Delta S^\circ$ is not affected by temperature in the same way. The positive entropy change that accompanies $\ce{A(g) -> B(g)}$, and therefore the increased number of possible microstates accessible to B isomers relative to A isomers remains the same at higher temperatures. Therefore, as $T\rightarrow\infty$, $K$ will no longer approach $K=1$, but will instead approach some value above 1, proportional to the magnitude of entropy increase.
Indeed, this is the reason why:
$$\lim_{T \to \infty} K
= lim_{T \to \infty} \exp(\frac{T\Delta S^\circ - \Delta H^\circ}{RT})
= \exp(\frac{\Delta S^\circ}{R})$$
The limit of $K$ as $T \rightarrow \infty$ is only dependent on $\Delta S^\circ$ and not on $\Delta H^\circ$.
Conclusion
In other words, we should, in a way, consider enthalpy and entropy as separately contributing to the equilibrium constant. Instead of saying: exothermicity means K decreases with increasing temperature and endothermicity means K increases with increasing temperature, we should see it as: at finite temperatures, exothermicity pushes the equilibrium constant toward $K>1$ and endothermicity pushes the equilibrium constant toward $K<1$, but with increasing temperature the influence of $\Delta H^\circ$ on $K$ diminishes. On the other hand, the influence of $\Delta S^\circ$ on the equilibrium constant does not wane with increasing temperature.
The contradiction is resolved by considering that the equilibrium constant is not solely dependent on the relative free energies of the reactants and products, but also on the entropy of the rest of the system, which is dependent on the temperature.
Of course, the above is highly simplified compared to real-life chemical reactions. Many assumptions were taken, but the goal was to help you appreciate, intuitively, why there is in fact no contradiction.
