Let's consider $SL(2,R)$ group, we know the following group element cannot be represented by exponential of the element of $sl(2,R)$, traceless real $2\times2$ matrices.
$$B=\begin{pmatrix} -a & 0\\ 0 & -1/a\end{pmatrix}$$ where $a>1$.
For any $A$ of $sl(2,R)$, the eigenvalues $\lambda_1$ and $\lambda_2$ of $A$ are either both real or both purely imaginary. In both cases $\lambda_1=-\lambda_2$. In the first case $e^{\lambda_1}$ and $e^{\lambda_2}$ are both real and positive, while in the second case $|e^{\lambda_1}|=|e^{\lambda_2}|=1$.
Suppose that $B=\exp(A)$ then $-a=e^{\lambda_1} $ which is impossible in this case.
In fact $$\exp[\begin{pmatrix} i\pi+\ln a & 0\\ 0 & i\pi-\ln a\end{pmatrix}]=B$$
So we see that $\begin{pmatrix} i\pi+\ln a & 0\\ 0 & i\pi-\ln a\end{pmatrix}$ do not belong to $sl(2,R)$.
However $B=\exp C.\exp D$, $$C=\begin{pmatrix} \ln a & 0\\ 0 & -\ln a\end{pmatrix}$$ and $$D= \begin{pmatrix} 0 & \pi\\ -\pi & 0\end{pmatrix}$$ where $C,D$ belong to $sl(2,R)$.
According to Baker–Campbell–Hausdorff formula, $\exp C. \exp D=\exp Z$ where $Z$ equals to the sum of $C,D$ and their commutators which still belongs to $sl(2,R)$. How to explain this paradox.
Wei, 1963 has, in essence your "counterexample", but you skipped the crucial part on convergence in the WP article you are referencing, a bit breezily.
You might rewrite your expressions to highlight the fact that your matrix D is $$ D=\pi E , $$ where $E^2=-1\!\!1$, so then, singularly, $$ \exp D= \cos \pi ~1\!\! 1+\sin \pi ~E = - 1\!\! 1 =\exp (i\pi 1\!\!1) ~. $$
D has norm π, so $\ln \pi^2 > \ln 2$ , violating the popular CBH convergence criterion pretty badly.