The exponential of a sum: commuting and non-commuting matrices

1. Commuting case. Since $AB=BA$, the binomial theorem gives
$$\frac{(A+B)^m}{m!}=\sum_{i+j=m}\frac{A^i}{i!}\cdot\frac{B^j}{j!}.$$
Summing over $m$ gives the Cauchy product of the series for $e^A$ and $e^B$. Both converge absolutely (problem on convergence of $\exp$), so the Cauchy product converges to the product of the sums:
$$e^{A+B}=\sum_m\sum_{i+j=m}\frac{A^i}{i!}\frac{B^j}{j!}=e^Ae^B.$$

2. A counterexample. Let $A=E_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $B=E_{21}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. Then $A^2=B^2=0$, so
$$e^A=I+A=\begin{pmatrix}1&1\\0&1\end{pmatrix},\qquad e^B=I+B=\begin{pmatrix}1&0\\1&1\end{pmatrix}.$$
Multiplying,
$$e^Ae^B=\begin{pmatrix}2&1\\1&1\end{pmatrix},\qquad e^Be^A=\begin{pmatrix}1&1\\1&2\end{pmatrix}.$$
These are different. Now $A+B$ is the symmetric matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, and $(A+B)^2=I$, so
$$e^{A+B}=\cosh1\,I+\sinh1\,(A+B)=\begin{pmatrix}\cosh1&\sinh1\\\sinh1&\cosh1\end{pmatrix}\approx\begin{pmatrix}1.543&1.175\\1.175&1.543\end{pmatrix},$$
which equals neither $e^Ae^B$ nor $e^Be^A$.

Remark. Since $e^Ae^B\ne e^Be^A$, the identity $e^{A+B}=e^Ae^B$ cannot hold when $A$ and $B$ do not commute: the left side is symmetric in $A,B$ but the right sides are not. The correct replacement is the Lie product formula (see below).