How can I tell if $A$ and $\exp(B)$ commute?
For $[A, B]$ it's simply $AB-BA$ and for $[\exp(A), \exp(B)]$ I think it'd be $\exp(A)\exp(B) - \exp(B)\exp(A) = \exp(A+B) - \exp(B+A) = 0$. Update: it's not generally true.
Is there a 'simple' way to find $[A, \exp(B)]$?
Or is this one of those problems where, if you encounter them at all, you are probably doing something wrong? The example I am encountering is $[\vec{S}, \exp(S_z)]$).
If OP wants to evaluate $[A,e^B]$ in terms of $[A,B]$, there is a formula
$$\tag{1} [A,e^B] ~=~\int_0^1 \! ds~ e^{(1-s)B} [A,B] e^{sB}. $$
Proof of eq.(1): The identity (1) follows by setting $t=1$ in the following identity
$$\tag{2} e^{-tB} [A,e^{tB}] ~=~ \int_0^t\!ds~e^{-sB}[A,B]e^{sB} .$$
To prove equation (2), first note that (2) is trivially true for $t=0$. Secondly, note that a differentiation wrt. $t$ on both sides of (2) produces the same expression
$$\tag{3} e^{-tB}[A,B]e^{tB},$$
where we use the fact that
$$\tag{4}\frac{d}{dt}e^{tB}~=~Be^{tB}~=~e^{tB}B.$$
So the two sides of eq.(2) must be equal.
Remark: See also this related Phys.SE post. (It is related because $[A, \cdot]$ acts as a linear derivation.)
$$[A,e^{B}]=[A,\sum_{i=0}^{\infty}\frac{B^{i}}{i!}]=\sum_{i=0}^{\infty}\frac{[A,B^{i}]}{i!}$$
so in order to $A$ and $e^B$ to commute, $A$ should commute with $B$ and hence with any power of $B$. You can apply this to $[\vec{S},e^{S_{z}}]$
$$[S_{i},e^{S_{z}}]=[S_{i},\sum_{j=0}^{\infty}\frac{S_{z}^{j}}{j!}]=\sum_{j=0}^{\infty}\frac{[S_{i},S_{z}^{j}]}{j!}$$
for $i=x,y,z$
