$\newcommand\si\sigma$Your condition on the moment generating function of $X$ implies that for such $|t|<1/b$ and $m=1,2,\dots$, $$\frac{t^{2m}}{(2m)!}\,E X^{2m}\le E\cosh tX\le\exp\frac{t^2 \si^2}{2(1-b|t|)}.$$ Taking now any real $c_1>0$ and any $c_2\in(0,1)$, and then choosing $t=\min(\frac{c_1}\si,\frac{c_2}b)$, we get $$E X^{2m}\le C(2m)!\max\Big(\frac\si{c_1},\frac{b}{c_2}\Big)^{2m},$$ with $C=\exp\frac{c_1^2}{2(1-c_2)}$. 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 C\sqrt{(2m)!(2m+2)!}\max\Big(\frac\si{c_1},\frac{b}{c_2}\Big)^{2m+1} \\ &\le \sqrt{\frac43}\, C(2m+1)!\max\Big(\frac\si{c_1},\frac{b}{c_2}\Big)^{2m+1}. \end{align*} So, we have the Bernstein-type condition \begin{align*} E|X|^k&\le\sqrt{\frac43}\, Ck!\max\Big(\frac\si{c_1},\frac{b}{c_2}\Big)^k \end{align*} for all $k\in\{2,3,\dots\}$.
One may note here that $\max\big(\frac\si{c_1},\frac{b}{c_2}\big)$ can be made arbitrarily close to $b$ by taking $c_2$ close to $1$ and then letting $c_1=c_2\si/b$. On the other hand, your condition on the moment generating function of $X$ is implied by the Bernstein condition $|EX^k|\le\frac{k!}2\,\si^2 b^{k-2}$ on moments of $X$ for $k=3,4,\dots$.
This is a modified/detailed version of the this previous answer.