Let $M$ be an $n\times n$ real matrix with strictly positive elements, i.e. $M_{ij}>0, \forall 1\leq i,j\leq n$, and let $\lambda_1$ be its Perron-Frobenius eigenvalue (i.e. the eigenvalue with largest magnitude, must be real and positive). Is the following inequality always correct? $$[e^{M}]_{ij}\leq e^{\lambda_1}, \forall 1\leq i,j \leq n$$
It is easy to prove this when $M$ is symmetric, since when $M$ is symmetric, we have $\|M\|=\lambda_1$, therefore $$[e^{M}]_{ij}\leq \|e^{M}\|\leq e^{\|M \|} = e^{\lambda_1}, \forall 1\leq i,j \leq n.$$ When $M$ is not symmetric, I checked several cases numerically and the answer seems to be true, but I don't know how to prove this. [The above proof fails because $\|M\|\neq \lambda_1$ when $M$ is not symmetric.]
This isn't true.
Suppose the claim was true:
In particular, consider positive stochastic matrix $M$ and
$\Sigma:= \begin{bmatrix}\sigma_{1}& \mathbf 0^T\\ \mathbf 0 &\mathbf I_{n-1} \end{bmatrix} $ with $\sigma_1 \gt 1$
$e^M$ is a positive matrix and has Perron Root $e$ thus
$\sigma_1\cdot\text{positive constant} = \big[\Sigma e^M \Sigma^{-1}\big]_{1,2} = \big[e^{\Sigma M\Sigma^{-1}}\big]_{1,2} \leq e$
but the upper bound is fixed while the lower bound may be made arbitrarily large by selecting large enough $\sigma_1$-- which is a contradiction.
working backwards, the counter example is
$M':= \Sigma M\Sigma^{-1}$
with sufficiently large $\sigma_1$