The Michaelis-Menten mechanism is the following:
\begin{align}
\ce{E + S &<-->[$k_1$][$k_{-1}$] ES} \tag{R1} \\
\ce{ES &->[$k_2$] P + E} \tag{R2}
\end{align}
If you apply: (1) steady-state approximation for the enzyme-substrate complex $\ce{ES}$, and (2) enzyme concentration balance $[E_\mathrm{t}] = [E] + [ES]$, then you arrive at the rate law or the Michaelis-Menten equation
\begin{equation}
r_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} =
\frac{k_2[E_\mathrm{t}] [S]}{[S] + K_\mathrm{M}} =
\frac{r_\mathrm{P,max} [S]}{[S] + K_\mathrm{M}} \tag{1}
\end{equation}
where $K_\mathrm{M} = (k_2 + k_{-1})/k_1$ and $r_\mathrm{P,max}:=k_2[E_\mathrm{t}]$, which is a constant.
[OP] I based the kinetics for these reactions on Michaelis-Menten kinetics, so every enzyme has a $V_{max}$ and $K_M$, and has velocity $v=V_{max}x /(K_M+x)$
The product $P$ has a maximum rate of $r_\mathrm{P,max}$, not the enzyme. This value is obtained when the substrate is so abundant that $[S] \gg K_\mathrm{M}$.
[OP] I use the Reaction Quotient $Q$ at any point in time to determine in which direction a reaction would proceed. E.g. would (net/overall) $A$ be converted to $B$ or vice versa. Each reaction is defined with a reaction energy which is used to calculate Equilibrium constant $K_e$.
You have as many reaction quotients $Q$ as reversible chemical reactions. You can do this for $\text{R1}$ because this step is reversible. However, you cannot do this for $\text{R2}$, because for the M-M mechanism this step is irreversible. Thus, once any quantity of $\ce{ES}$ complex has been formed, $\text{R2}$ will proceed, independently if you had a lot of $\pu{P}$ at $t=0$.
My understanding is that Miachelis-Menten kinetics only describe the case in which $Q<<K_e$. What happens if $Q→K_e$?
A reaction quotient is not defined for irreversible reactions.
Intuitively I would assume that the overall reaction velocity slows down as $Q$ approaches $K_e$. Is there a term that I can add to the Michaelis-Menten equation which would describe this decreasing $v$?
I propose to modify the reaction mechanism in the second step
\begin{align}
\ce{ES &<-->[$k_2$][$k_{-2}$]P + E} \tag{new R2}
\end{align}
However, the math has to be carried again, because the old $\text{R2}$ had a rate of $k_2[ES]$, and now we have
\begin{align}
r_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} =
k_2[ES] - k_{-2} [P][E] \tag{2}
\end{align}
This adds an additional rate constant $k_{-2}$ to your simulation. Another option is to consider the Briggs-Haldane rate law, where every step is reversible. If you need help with the math just tell in the comments.