Is commutation preserved for $e^A$?

Question

Let $A, B \in \mathbb C^{n \times n}$. Suppose $A$ commutes with $B$. Does $e^A$ necessarily commute with $B$?

If that is the case, consider $$S = A + B \exp(-S)$$ where $S$ is an $n \times n$ matrix. Does $S$ necessarily commute with $A$?

Answer 1

More generally, for any analytic function $f$ on a neighbourhood of the spectrum of $A$, $f(A)$ (as defined by the holomorphic functional calculus) is a limit of rational functions of $A$, and therefore commutes with $B$.

For your second question, formally a solution is $S = A + W(B \exp(-A))$ where $W$ is a branch of the Lambert W function. The various branches of $W$ have various branch points, but if I'm not mistaken, for every point of the complex plane there is a branch of $W$ that is analytic there. Thus it should always be possible to produce a branch that is analytic in a neighbourhood of the spectrum of $B \exp(-A)$, and thus to produce a corresponding solution $S$. Then since $B$ commutes with $A$, so does $S$.
I'm not sure if every possible solution $S$ must commute with $A$.