I know it is true in $\mbox{SL}_2(\Bbb C)$ and I think it is true in $M_n(\Bbb C)$ because if $M=PDP^{-1}$, we might be able to write D as $\exp(E)$ for some $E\in M_n(\Bbb C)$ as the exponential is surjective from $\Bbb C$ onto $\Bbb C^*$ and all eigenvalues of M are non zero because they are distinct.
Thank you for your help.
A diagonalizable matrix is an exponential (over $\mathbb C$) if and only if it is not a singular matrix. Of course, no singular matrix can be an exponential, since $\det e^A=e^{\operatorname{tr}A}\neq0$. On the other hand, if $A$ is diagonalizable, then it is similar to a diagonal matrix$$\begin{bmatrix}d_1&0&0&\ldots&0\\0&d_2&0&\ldots&0\\0&0&d_3&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&d_n\end{bmatrix}.$$Then $A$ is non-singular if and only if every $d_k$ is non--zero. So, let $\lambda_k$ be a logarithm of $d_k$ and$$\exp\left(\begin{bmatrix}\lambda_1&0&0&\ldots&0\\0&\lambda_2&0&\ldots&0\\0&0&\lambda_3&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&\lambda_n\end{bmatrix}\right)=\begin{bmatrix}d_1&0&0&\ldots&0\\0&d_2&0&\ldots&0\\0&0&d_3&\ldots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\ldots&d_n\end{bmatrix}.$$So, $A$ is exponential.