Find square matrices $A, B$, such that $\exp(A + B) \neq \exp(A) \exp(B)$. The question is as the Title stated:
I picked a very easy example. However, I am afraid, I am missing something.
The two matrices that I picked are 
$$A = \begin{bmatrix}1& 0 \\ 0 & 0\end{bmatrix}\quad \text{and}\quad 
B =\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$
My reasoning is that 
$$
\exp(A) = \begin{bmatrix}e^1& 0 \\ 0 & 0\end{bmatrix}\quad\text{and}\quad
\exp(B) =\begin{bmatrix}0 & 0 \\ 0 & e^1\end{bmatrix}$$
$$\exp(A)\exp(B) = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}$$
and $$exp(A+B) = \begin{bmatrix} e^1 & 0 \\ 0 & e^1\end{bmatrix}.$$
Please check it for me
 A: In your example
$$A = \begin{bmatrix}0& 0 \\ 1 & 0\end{bmatrix}\quad \text{and}\quad 
B =\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$
we have that
$A^2=A^k=0$, for all $k\ge 2$, and $B^k=B$, for all $k\ge 1$, and therefore
$$
\exp A=I +A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!}+\cdots=I+A=
\begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix},
$$ 
while
$$
\exp B=I +B+\frac{B^2}{2!}+\cdots+\frac{B^n}{n!}+\cdots=I+B+\frac{B}{2!}+\cdots+\frac{B}{n!}+
\cdots=I+(\mathrm{e}-1)B
=\begin{bmatrix}1 & 0 \\ 0 & \mathrm{e}\end{bmatrix}.$$
and
$$\exp(A)\exp(B) = \begin{bmatrix}1 & 0 \\ 1 & \mathrm{e}\end{bmatrix}\quad\text{and}\quad
\exp(A+B) = \begin{bmatrix}1 & 0 \\ \mathrm{e}-1 & \mathrm{e}\end{bmatrix}.$$
Thus your example WORKS!
Note. A characterization of the commutativity of the exponential of matrices is provided by:
Theorem. Let $A,B\in\mathbb R^{n\times n}$. Then
$$
\mathrm{e}^{tA}\mathrm{e}^{tB}=\mathrm{e}^{t(A+B)},
$$
for all $t\in\mathbb R$, if and only if $AB=BA$.
For a proof Theorem 5 in these notes.
