For two non-commuting matrices $A,B \in M(2,\mathbb{K})$, $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$, can be shown that: $$ e^C=e^{A+B}=e^Ae^B=e^Be^A \iff \begin{cases} \mbox{det}A'=\mbox{det}A-\left(\dfrac{\mbox{tr}A}{2}\right)^2=m^2\pi^2\\ \mbox{det}B'=\mbox{det}B-\left(\dfrac{\mbox{tr}B}{2}\right)^2=n^2\pi^2\\ \mbox{det}C'=\mbox{det}C-\left(\dfrac{\mbox{tr}C}{2}\right)^2=q^2\pi^2\\ \end {cases} $$ with: $m,n,q \in \mathbb{N}^*$ and $(-1)^m (-1)^n=(-1)^q$.
Here a sketch of the proof:
1) Note that we can always put a $2 \times 2$ matrix in the form $A=hI+A'$ where: $h=\mbox{tr}A/2$, $\mbox{tr}A'=0$, $\mbox{det}A'=\mbox{det}A-\left(\dfrac{\mbox{tr}A}{2}\right)^2$; and $[hI,A']=0$ so that $e^A=e^he^{A'}$.
2)Using the series expansion we see that, for a traceless matrix, we have: $$e^{A'}= {I}\,\cos\alpha+A'\;\dfrac{\sin\alpha}{\alpha}$$ where $\alpha=\sqrt{\mbox{det}A'}$.
3) Using this result for non-commuting $A,B$ we have: $$[e^{A},e^{B}]=e^{(h+k)}\dfrac{\sin \alpha \sin \beta}{\alpha \beta}[A',B']$$ (where $k=\mbox{tr}B/2$ and $\beta=\sqrt{\mbox{det}B'}$) and, since $[A',B']=[A,B] \ne 0$, for $[e^{A},e^{B}]=0$ must be $\alpha=m\pi$ or $\beta=n \pi$ with $m,n \in \mathbb{N}^*$.
4)So we set $\alpha=m\pi$ and calculate $e^{A+B}$ and $e^A e^B$. Equating the corrisponding elements of the two matrices and solving the system we have the assertion.
This proof does not extend to matrices $n \times n$ and the best result I could find in the literature is a sufficient condition [Edgar M. E. Wermuth: A REMARK ON COMMUTING OPERATOR EXPONENTIALS] that says:
Let A and B be bounded operators on a Banach space with $2\pi i$-congruence-free spectra. Then $e^Ae^B = e^B e^A$ if and only if $AB = BA$. Where a set $\Lambda$ of complex numbers is $2\pi i$-congruence-free, if there are no two different elements $\lambda_1 , \lambda_2 \in \Lambda$ such that $\lambda_1 \equiv \lambda_2$ (mod $2\pi i$). (Clearly the previous result in $M(2,\mathbb{K})$ is consistent with this statement).
I'd like to know if anyone knows strongest results on this topic for $n \times n$ matrices.