$\newcommand\si\sigma$After the clarification by the OP, my previous answer should be modified as follows.
Suppose that for some real $C$, some real $h>0$, and all $t\in[-h,h]$ we have $$M(t):=Ee^{tX}\le C \tag{1}\label{1} $$ (note that necessarily $C\ge M(0)=1$.) We will show that then $$Ee^{tX}\le1+\frac{\si^2 t^2/2}{1-b|t|}\le\exp\frac{\si^2 t^2/2}{1-b|t|} \tag{2}\label{2} $$ for $$b:=\frac{4C}{\sqrt3\,h^3\si^2} \tag{3}\label{3} $$ and all $t\in[-1/b,1/b]$ (excluding the trivial case $\si=0$). Note that $C\ge M(h)\ge 1+\si^2 h^2/2>\si^2 h^2/2$ and hence $b h>\frac2{\sqrt3}>1$, so that $b>1/h$.
Indeed, \eqref{1} implies that for $m=1,2,\dots$ $$\frac{h^{2m}}{(2m)!}\,E X^{2m}\le E\cosh hX\le C,$$ so that $$E X^{2m}\le C\frac{(2m)!}{h^{2m}}.$$ Using now the Cauchy--Schwarz inequality, for $m=1,2,\dots$ we get \begin{align*} E|X|^{2m+1}&\le\sqrt{E X^{2m}\,E X^{2m+2}} \\ &\le CC\frac{\sqrt{(2m)!(2m+2)!}}{h^{2m+1}} \\ &\le \sqrt{\frac43}\, C\frac{(2m+1)!}{h^{2m+1}}. \end{align*} So, \begin{align*} E|X|^k&\le\sqrt{\frac43}\, C\frac{k!}{h^k} \end{align*} for all $k\in\{3,4\dots\}$.
So, for $b$ as in \eqref{3} and $t\in(-1/b,1/b)$, $$Ee^{tX}=\sum_{k=0}^\infty EX^k\frac{t^k}{k!} \le1+\frac{\si^2 t^2}2+\sum_{k=3}^\infty \sqrt{\frac43}\, C\frac{k!}{h^k}\frac{|t|^k}{k!} \\ =1+\frac{\si^2 t^2}2+\sqrt{\frac43}\, C\frac{(|t|/h)^3}{1-|t|/h} \le1+\frac{\si^2 t^2/2}{1-b|t|},$$$$Ee^{tX}=\sum_{k=0}^\infty EX^k\frac{t^k}{k!} \le1+\frac{\si^2 t^2}2+\sum_{k=3}^\infty \sqrt{\frac43}\, C\frac{k!}{h^k}\frac{|t|^k}{k!} \\ =1+\frac{\si^2 t^2}2+\sqrt{\frac43}\, C\frac{(|t|/h)^3}{1-|t|/h} \le1+\frac{\si^2 t^2/2}{1-b|t|}, \tag{4}\label{4} $$ so that \eqref{2} is proved. $\quad\Box$
One may also note that the latter inequality in \eqref{4} is exact in the sense that $$1+\frac{\si^2 t^2}2+\sqrt{\frac43}\, C\frac{(|t|/h)^3}{1-|t|/h} =1+\frac{\si^2 t^2/2}{1-b|t|}-O(t^4)$$ for $t\downarrow0$.