Let $D$ be a matrix having positive off-diagonal values and nonpositive diagonal values such that the row sums are nonpositive and $D$ is invertible. Then $-D$ is an M-matrix. Now decrease some diagonal entry $D_{ii}$, i.e. let $E$ be a diagonal matrix with $E_{ii} < 0$ and $E_{jj} = 0$ for $j \neq i$ and consider the matrix $D + E$. Is it true that $$ e_k \text{exp}(D + E) e \leq e_k \text{exp}(D) e $$ where $e_k$ is a unit vector and $e$ a vector full of ones. Maybe it is more true, namely that $$ \text{exp}(D + E) \leq \text{exp}(D) $$ componentwise.
In general, $D$ and $E$ do not commute. Possibly there is some trick that does not appeal to the Zassenhaus formula.
For $A_1, A_2 \in \mathbb{R}^{n \times n}$ denote $A_1 \leq A_2$ whenever the entries of $A_1$ are less or equal than the corresponding entries of $A_2$.
First of all, if $0 \leq A_1 \leq A_2$ then clearly $A_1^k \leq A_2^k$ for all $k \in \mathbb{N}$ and thus $e^{A_1} \leq e^{A_2}$.
For the matrix $D + E$ above there exists $a > 0$ and a matrix $A_1 \geq 0$ such that $D + E = A_1 - a I$. It follows that $D = A_2 - a I$ with $A_2 := A_1 - E$ and it holds that $0 \leq A_1 = A_2 + E \leq A_2$ since $E \leq 0$. Therefore,
$$ e^{D+E} = e^{A_1} e^{-a} \leq e^{A_2} e^{-a} = e^D. $$