On equation $e^{xy-yx}=e^xe^ye^{-x}e^{-y}$ in $C^*$ algebras Inspired by this MSE  question we ask the following question:
Is there a noncommutative $C^*$-algebra $A$ for which the following identity holds for all $x,y \in A$?
$$e^{(xy-yx)}= e^xe^y e^{-x}e^{-y}$$
That is $$e^{[x,y]}=[e^x,e^y]$$
where the bracket on the left-hand side is the algebra commutator, and the bracket on the right-hand side denotes the group commutator.
 A: Yes:

A $C^*$-algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ iff it is commutative. 

This follows from two independent facts (I write $[x,y]=xy-yx$)

1) A (real/complex) unital Banach algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ $\Leftrightarrow$ it satisfies the identity $[x,[x,y]]=0$ $\Leftrightarrow$ it satisfies the identity $[x,[y,z]]$, i.e., its underlying Lie algebra is 2-step nilpotent.
2) A $C^*$-algebra satisfying the identity $[x,[x,y]]=0$ is commutative.

For (1), one obtains that the main implication, namely the identity implies $[x,[x,y]]=0$, by a 3rd order Taylor expansion: in every Banach algebra one has 
$$e^{xy-yx}=1+xy-yx+o(\|x\|^3+\|y\|^3);$$
$$e^xe^ye^{-x}e^{-y}=1+xy-yx+\frac12([x+y,[x,y]])+o(\|x\|^3+\|y\|^3),$$
so the commutator identity forces $[x+y,[x,y]]=0$ for all $x,y$, and hence $[x,[x,y]$ identically vanishes by homogeneity.
(As mentioned in the comments, a simple argument then implies $[x,[y,z]]=0$ for all $x,y,z$, but this is not needed to run the argument.)
Conversely, if $[x,[x,y]]$ is identically zero, the Baker-Campbell-Hausdorff formula reads as: for all $x,y$, one has $\exp(x)\exp(y)=\exp(x+y+(1/2)[x,y])$. The commutator identity follows.
Let us pass to (2). Let $A$ be a $C^*$-algebra satisfying the PI-identity $[x,[x,y]]=0$. First, if $A$ is finite-dimensional, it is a product of matrix algebras, and hence is commutative (since the given identity fails for $M_n(\mathbf{C})$ for $n\ge 2$). In general, I need the last emphasized statement in my answer here: $A$ has a family $(J_i)$ of finite-codimensional 2-sided closed $*$-ideals with trivial intersection (so $A$ embeds into the product $\prod_i A/J_i$). By the finite-dimensional case, $A/J_i$ is commutative for every $i$. Hence $A$ is commutative.
