Suppose you have an ideal gas in a container in mechanical and thermal equilibrium with its surroundings (same $T$ and $p$—say, 298 K and 1 atm). It can't do any work on the surroundings.
Now compress it isothermally to, say, 100 atm. Now it certainly can do expansion work on the surroundings. What change in thermodynamic state function accounts for its increased ability to do work?
Well, since the internal energy, $U$, of a sample of ideal gas depends solely on its temperature, its internal energy hasn't changed. So the additional capacity to do work has not resulted from an increase in $U$.
But its free energy is much higher. Since $\Delta U = 0$ for that isothermal compression, the increase in the system's free energy—and the corresponding increase in the its ability to do work—has resulted solely from a decrease in its entropy.
So yes, a change in entropy can be used to do work.
It's tricky to talk formally about free energies for the above system, since while it's constant $T$, it's neither constant $p$ nor constant $V$. Rather it's constant $p \times V$.
So let's consider some more easily characterized systems (ideal gases at constant $T$ and $V$ or constant $T$ and $p$):
In a closed system at constant $T$ and $V$, the maximum total work the system can do, which is the reversible work, is given by $\Delta A$, the Helmholtz free energy. Under those conditions, $\Delta A=\Delta U - T\Delta S$. If $\Delta U = 0$, the magnitude of the maximum work is given by $T\Delta S$.
And in a closed system at constant $T$ and $p$, the maximum total non-$pV$ work the system can do, which is the reversible non-$pV$ work, is given by $\Delta G$, the Gibbs free energy. Under those conditions, $\Delta G = \Delta H - T\Delta S$. If $\Delta H= 0$, the magnitude of the maximum non-$pV$ work is given by $T\Delta S$.
So you can see we can formally connect the work a system can do with the entropy change. This shouldn't be surprising, because nothing happens to a macroscopic system unless it's associated with an increase in the entropy of the universe. That's the 2nd law of thermodynamics, which provides an arrow of time for the universe. Thus anytime work is done anywhere, it's always ultimately driven by an increase in entropy (not of the system, but of the universe).
An increase in the entropy of the universe is the driving force behind every macroscopic process.
At constant $T$ and $p$, the only reason a decrease in enthalpy (release of thermal energy) corresponds to the ability of a system to undergo a change is because it causes an increase in the entropy of the surroundings: $\Delta H_{sys} = -T\Delta S_{surr}$. Thus ultimately the abilty to undergo any process, including those in which the system does work, isn't about the change in the energy of the system. It's about the change in the entropy of the universe.
Let's make this more general (i.e., not restrict our discussion to ideal gases):
In a closed system at at constant $T$, $V$, both of the following hold (I'm taking positive $đw$ to be work done on the system):
$$dA = -T(dS_{sys}+dS_{surr}) = -T dS_{univ}$$
$$dA = đw_{max}$$
Thus:
$$đw_{max} = -T dS_{univ}$$
And in a closed system at at constant $T$, $p$:
$$dG = -T(dS_{sys}+dS_{surr}) = -T dS_{univ}$$
$$dG = đw_{non-pV, max}$$
[G only gives the non-$pV$-work because, at constant $p$, the $pV$-work is subtracted from G because it's subtracted from the enthalpy: $H = U +pV$; at constant $p$: $dH = dU + pdV = dU - đw_{pV}$.]
Thus:
$$đw_{non-pV,max} = -T dS_{univ}$$
In sum, it's all about the entropy, baby! The change in the energy or enthalpy of the system is only a bookkeeping surrogate for what really matters—the change in the entropy of the universe (unless you're doing industrial chemistry, and need to know the heat flow for practical reasons, in which case you would want to know $\Delta H$ or $\Delta U$, for processes at constant $p$ or $V$, respectively).
