Obviously, if
$$\mathbf{A} = \begin{bmatrix}\mathbf{C} & \mathbf{0}\\ \mathbf{0} & \mathbf{D}\end{bmatrix}$$
then
$$e^{A}=\begin{bmatrix}\mathbf{e^C} & \mathbf{0}\\ \mathbf{0} & \mathbf{e^D}\end{bmatrix}$$
Given any invertible matrix $X$ we can express it as $X = e^A$ (but complex matrix $A$ is not unique until $X$ is not from one-parameter subgroup of $\mbox{GL}(n, \mathbb{C})$). Anyway, given that $X$ has a block diagonal structure then does $A$ need to have block diagonal structure as well?
No. A counterexample is $\mathbf A=\pmatrix{0&2\pi\mathrm i\\2\pi\mathrm i&0}$, with $\mathrm e^{\mathbf A}=I$.