If $|A|, |B|$ are sufficiently small, does $e^{A+B} = e^A e^B$ imply that $AB = BA$? For  complex matrices $A, B$ of the same size, there are simple counterexamples to the question,
"Does $e^{A+B} = e^A e^B$ imply that $AB = BA$?''; but they do not involve matrices that have very small
(Euclidean) norm. See for example, one of the responses to this question, in which $e^{A+B} = e^A = e^B = I$, where $I$ is the identity matrix, but $A$ and $B$ do not commute.
If there are counterexamples with $|A|, |B|$ being arbitrarily small then that settles the question. 
In the alternative, here is the small progress that I have made, in case the following line of reasoning can
be continued to a solution.
First, by "very small'', I mean that the Campbell-Baker-Hausdorff (CBH) series expansion is valid
for both $\log(e^A e^B)$ and $\log(e^B e^A)$. The CBH series expansion for $\log(e^X e^Y)$ is 
$e^X e^Y = \exp( \sum_{m=1}^{\infty} F_m(X, Y) )$, where $F_m(X, Y)$ is a homogeneous polynomial
of degree $m$ in the (generally, non-commuting) variables $X, Y$ (complex matrices of equal size, $n$, say); 
and $F_m(X, Y)$ contains neither the term $X^m$ nor the term $Y^m$, except for $m=1$: $F_1(X, Y) = X + Y$.
Each $F_m(X, Y)$ is moreover a linear combination of nested Lie brackets of $X$ and $Y$. Also, we have the
elementary property, $F_m(Y, X) = (-1)^{m+1} F_m(X, Y)$.
Next, it suffices to show that $e^{A+B} = e^A e^B$ implies that $e^{A+B} = e^B e^A$, for then
$e^B e^A = e^B e^A$, which does imply that $AB = BA$, under the assumption that $|A|, |B|$ are sufficiently small.
Indeed, setting $X = e^A$ and $Y = e^B$, so that $A = \log X$ and $B = \log Y$, then $AB = BA$ because
$\qquad A = (X−I) − \frac{1}{2}(X−I)^2 + \frac{1}{3}(X−I)^3 − \cdots$,
$\qquad B = (Y−I) − \frac{1}{2}(Y−I)^2 + \frac{1}{3}(Y−I)^3 − \cdots$,
and the series commute since $X$ and $Y$ do.
The assumed equality, $e^A e^B = e^{A+B}$, implies that $\sum_{m=2}^{\infty} F_m(A, B) = 0$; but $A$ and $B$
are specific matrices, so we can't (immediately) conclude that each $F_m(A, B) = 0$, for $m \geq 2$. 
Using the property, $F_m(B, A) = (-1)^{m+1} F_m(A, B)$, we have
$$
 e^B e^A = \exp\left( \sum_{m=1}^{\infty} (-1)^{m+1} F_m(A, B) \right) = \exp\left( X + Y + \sum_{m=2}^{\infty} (-1)^{m+1} F_m(A, B) \right).
$$
Based only on the equality, $\sum_{m=2}^{\infty} F_m(A, B) = 0$, I don't see why $\sum_{m=2}^{\infty} (-1)^{m+1} F_m(A, B)$
should equal $0$.
Another approach might be to consider $e^A e^B e^{-A} e^{-B}$.
 A: The OP's question seems to be:
Does there exist $a>0$ s.t. if $A,B\in M_n(\mathbb{C})$ satisfy $(*)$ $e^{A+B}=e^Ae^B$ and $||A||<a,||B||<a$, then $AB=BA$ ?
I think that it is a very difficult question. When $n\geq 3$, nobody knows how to solve $(*)$. Yet, when $n=2$, we know all the non-commuting couples solutions of $(*)$.
$\textbf{Proposition 1}$. When $n=2$, a positive answer is given by $a=\pi/2$.
$\textbf{Proof}$. Let $U=\{u\in\mathbb{C}^*;e^u=1+u\}$. Note that, for every $u\in U$, $||u||>7$.
We assume that $(*)$ is satisfied and $AB\not= BA$. Then there are $3$ cases
Case 1. For example, $e^A$ is a homothety and there are $\lambda\in\mathbb{C}$ and $l\in\mathbb{Z}^*$ s.t. $spectrum(A)=\{\lambda+i\pi l,\lambda-i\pi l\}$. Then $\rho(A)\geq \pi$ and, for every operator norm, $||A||\geq \pi$.
Case 2. For example, $A=\lambda I_2+A'$ where $\lambda\in\mathbb{C}$ and $e^{A'}=I_2+A'$; there is $u\in U$ s.t. $spectrum(A)=\{\lambda+u,\lambda\}$. Then $\rho(A)\geq ||u||/2>\pi$.
Case 3. Up to additive homotheties, $A,B$ are simultaneously similar to $diag(0,u),\begin{pmatrix}v&1\\0&0\end{pmatrix}$, where $u,v\in\mathbb{C}^*,u\not=v$ and $\dfrac{e^u-1}{u }=\dfrac{e^v-1}{v}\not= 0$.
Assume that $||A||<\pi/2,||B||<\pi/2$. Then $||u||/2\leq \rho(A)<\pi/2$, $||u||<\pi$ and $|Im(u)|<\pi$ ($Im(u)$ denotes the imaginary part of $u$); in the same way $|Im(v)|<\pi$. One can prove that necessarily (it's not obvious) $|Im(u)-Im(v)|\geq 2\pi$, a contradiction. $\square$
The equation $e^Ae^B=e^Be^A$  is easier to study. One has the following
$\textbf{Proposition 2}$. If $A,B\in M_n(\mathbb{C})$ satisfy $e^Be^A=e^Ae^B$ and $||A||<\pi,||B||<\pi$, then $AB=BA$.
$\textbf{Proof}$. Definition. We say that a complex finite subset $((a_i))$ is $2i\pi$ congruence-free (denoted $2i\pi$ CF) iff there are no distinct $a_i,a_j$ s.t. $a_i-a_j\in 2i\pi\mathbb{Z}^*$.
Let $A,B\in M_n(\mathbb{C})$ s.t. $e^Be^A=e^Ae^B$ and let
$spectrum(A)=(\lambda_i)$, $spectrum(B)=(\mu_i)$. Assume that $AB\not= BA$. Then the spectra of $A,B$ are not both $2i\pi$ CF, cf.
http://www.ams.org/journals/proc/1997-125-06/S0002-9939-97-03643-5/S0002-9939-97-03643-5.pdf
For example, there are $i\not= j$ s.t. $\lambda_i-\lambda_j\in 2i\pi\mathbb{Z}^*$. Thus $\rho(A)\geq \pi$ and, for every operator norm, $||A||\geq \pi$.  $\square$
