$\newcommand\si\sigma$After the [clarification by the OP][1], my [previous answer][2] 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 
$$b:=\max\Big(h,\frac{4C}{\sqrt3\,h^3\si^2}\Big) \tag{3}\label{3} $$
and all $t\in[-1/b,1/b]$. 

Indeed, \eqref{1} implies that for $m=1,2,\dots$ 
$$\frac{h^{2m}}{(2m)!}\,E X^{2m}\le E\cosh tX\le C,$$
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 $t\in(-1/b,1/b)$, 
$$Ee^{tX}=\sum_{k=0}^\infty EX^k\frac{t^k}{k!}
\le1+\frac{\si^2 t^2}2+\sum_{k=3}^\infty \sqrt{\frac43}\, C\frac{k!}{h^k}\frac{|t|^k}{k!} \\ 
=1+\frac{\si^2 t^2}2+\sqrt{\frac43}\, C\frac{(|t|/h)^3}{1-|t|/h}
\le1+\frac{\si^2 t^2/2}{1-b|t|},$$
so that \eqref{2} is proved. $\quad\Box$


  [1]: https://mathoverflow.net/questions/481701/existence-and-sharpness-of-bernstein-type-bounds-on-the-moment-generating-functi/481703#comment1254369_481701
  [2]: https://mathoverflow.net/a/481703/36721
  [3]: https://mathoverflow.net/a/479804/36721