Commutator with exponential $[A, \exp(B)]$

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 $i=x,y,z$