Apparently your problem is only $A x = y$,
$F \circ x \preceq g$ it could make your problem infeasible, I wouldn't artificially add it. If you don't know what $F$ or $g$ to use, you could start with $F=0$, and $g=0$, so that accepts any $x$, in other words don't use that constraint.
About the entropy maximization, you drop a negative sign in your expression, you must maximize $\sum -x \log(x)$
The regularization effect is related to convexity, the derivative of $-x \log(x)$ is $-(1 + \log(x))$. As $x$ increase the derivative increase, so if you have degrees of freedom that let you vary $x_i$ and $x_j$, the entropy maximization will pull those two values closer to each other, i.e. you can increase entropy by increasing the smaller and decreasing the greater of them.
In particular, a well known result is that under the constraint $\sum x=1$, the uniform distribution maximizes entropy.
Summarizing
The L1 regularization will pull values towards zero. Maximum entropy regularization will pull the values closer to each other.
Edit
How to use it for positive and negative values.
If you want to use this regularization for positive and negative values you could think of some transformation $\mathbb{R} \to \mathbb{R}^+$.
If you use $\exp(x)$ as you mentioned in the comment, you end up maximizing $ \sum x \exp(x)$. Negative values will have negligible contributions, while positive values may have huge contributions to the sum, so it will probably move the negative values to as a function of the positive ones.
Also,the minimum is $0 = x \exp(x) + \exp(x) = (x+1) \exp(x)$, what $x=1", has in special.
Maybe it would be preferrable to use $\exp(x^2)$, then the positive and negative side will look the same. The critical points are $(2x \cdot x^2 + 2x)\exp(x^2) =2\, x \,(x^2+1) \exp(x^2) = 0$, i.e. it has a minimum in $x=0$.
The other point is that $\exp(x)$ or $\exp(x^2)$ is prone to overflow. If you want to avoid problems due to the exponential growth of the exponential function you could try $x^2$, but this is not convex any more $x + 2x \log(x^2)=0$, has three solutions $x=0$, $x=\pm e^{-1/4}$.
Then the things starts to get too complicated, and this is not entropy maximization anymore. If you are open to other regularizations I would give L2 regularization a chance.
