Prove that $e^{t(X+Y)}=e^{tX} e^{tY}$ implies $[X,Y]=0$

Take $ A=\begin{pmatrix} 0&0 \\ 0&2i\pi \\ \end{pmatrix}$, $\quad B=\begin{pmatrix} 0&1\\ 0 & 2i\pi \end{pmatrix}.$ Then you can show that $\exp(A)=\exp(B)=\exp(A+B)=I_2$, and $AB \neq BA$.

For the proof of $e^{Xt+Yt}=e^{Xt} e^{Yt} \Rightarrow [X,Y]=0$ : note that $$\exp(\frac{Xt}{n}).\exp(\frac{Yt}{n}).\exp(-\frac{Xt+Yt}{n}) = {\rm Id} + \frac{[Xt,Yt]}{2n^2} + o(n^{-2}),$$ hence $$\lim_{n \to +\infty} \left( \exp(\frac{Xt}{n}).\exp(\frac{Yt}{n}).\exp(-\frac{Xt+Yt}{n}) \right)^{2n^2} = \exp([Xt,Yt]).$$ since the sequence is constant to $1$, we have $\exp([Xt,Yt])=1$. Take the derivative at $t=0$ to conclude that $[X,Y]=0$.