Suppose we know $\mathbf{E} e^{sX} \leq M(s)$. Can we use $M(s)$ to get bounds on individual moments $m_n = \mathbf{E}X^n$?
The other direction holds, since if $m_n \leq M_n$ for all $m=0,1,2,\ldots$ then $$\mathbf{E} e^{sX} = \sum_{n=0}^\infty \frac{s^n}{n!}\mathbf{E} X^n = \sum_{n=0}^\infty \frac{s^n}{n!}m_n \leq \sum_{n=0}^\infty \frac{s^n}{n!} M_n = M(s)$$
Suppose you have $\mathbb{E}\exp(sX)\leq M(s)$ for all $s\in (-\delta,\delta)$. Then by DCT, $\left\lvert\mathbb{E}\exp(zX)\right\rvert\leq M(\operatorname{Re} z)$ for all $z$ in the strip, and $\mathbb{E}\exp(zX)$ is analytic there, hence we recover $$ \mathbb{E}X^n=\left.\frac{\mathrm{d}^n}{\mathrm{d}z^n}\right\vert_{z=0}\mathbb{E}\exp(zX)=\frac{n!}{2\pi i}\int_C\frac{\mathbb{E}\exp(zX)}{z^{n+1}}\,\mathrm{d}z $$ where $C$ is a small circular contour around $0$ inside the strip. So the usual M-L estimate gives $$ \mathbb{E}X^n\leq n!r^{-n}\sup \{M(s):s\in [-r,r]\} $$ or a sharper bound using $\left\lvert\int_I f\right\rvert\leq\int_I\lvert f\rvert$ instead: $$ \mathbb{E}X^n\leq n!r^{-n}\int_0^1 M(r\cos 2\pi t)\,\mathrm{d}t. $$ where $r<\delta$.