Suppose $e^A = A$, prove that $A$ is diagonalizable

You write $$D+N=e^{D}e^{N}=e^{D}+e^{D}(e^{N}-I)$$

As the matrix $e^{D}(e^{N}-I)$ is nilpotent (because it is of the form $e^{D}NQ(N)$ and everything commutes), $e^D$ is diagonalizable (because $D$ is), and these two matrices commutes, from uniqueness in Dunford decomposition, you get:

$$D=e^D, \;\;\;\;N=e^{D}(e^{N}-I)$$

Therefore you get, multiplying by $N^{k-1}$ where $k$ is the smallest integer such that $N^{k+1}=0$ (and assuming by contradiction $k\geq 1$):$$N^k=DN^k,$$ in other words $$(I-D)N^k=0.$$ Now, is $(I-D)$ inversible? Yes, as $D=e^D$, 1 cannot be an eigenvalue for $D$, and you conclude $N^k=0$, which by definition of $k$, constitute a contradiction, and finish the proof as $k=0$ and $N=0$.

It is sufficient to look at a Jordan block $J$ of $A$. Such a block, of size $r\geq1$, has an eigenvalue $\lambda$ of $A$ along its main diagonal, ones in the upper secondary diagonal, and otherwise zeros: $$J=\lambda I+N\ .$$ As $\lambda I$ commutes with $N$ the condition $J=e^J$ leads to $$\lambda I+N=e^{\lambda I}\cdot e^N=e^\lambda\left(I+N+{N^2\over 2!}+{N^3\over 3!}+\ldots\right)\ .\tag{1}$$ Looking at the diagonal elements here we see that $\lambda\in{\mathbb C}$ has to satisfy $$\lambda=e^\lambda \ .\tag{2}$$ Now we look at the secondary diagonal in $(1)$ and find that $\lambda$ has to satisfy $1=e^\lambda \cdot 1$ as well, unless $r=1$. The equation $(2)$ has an infinity of solutions, but none of them is $=2k\pi i$ with integer $k$. It follows that in fact $r=1$, and this implies that all Jordan blocks of $A$ have size $1$, hence $A$ is diagonalizable.