Asymptotics of $\sum\limits_{n=2}^\infty \frac{x^n}{(\log n) n!}$ as $x\to\infty$

Consider the related finite sum $$ s(x)=\sum\limits_n\frac{x^n}{n!}\mathbf 1_{x/2\leqslant n\leqslant 2x}=\mathrm e^x\cdot\mathbb P(x/2\leqslant N_x\leqslant 2x), $$ where $N_x$ is a Poisson random variable with parameter $x$. A large deviation estimates ensures that, for every $x$ large enough, $\mathbb P(x/2\leqslant N_x\leqslant 2x)=1-p(x)$ where $0\lt p(x)\leqslant\mathrm e^{-cx}$ for some positive $c$.

The series $f(x)$ to be estimated is $f(x)=g(x)+h(x)$ where $g(x)$ sums the terms such that $x/2\leqslant n\leqslant2x$ and $h(x)$ sums the other terms. Note that $0\leqslant h(x)\leqslant\mathrm e^x p(x)/\log2$ and $s(x)/\log(2x)\leqslant g(x)\leqslant s(x)/\log(x/2)$, hence $$ \mathrm e^{x}\frac{1-\mathrm e^{-cx}}{\log x+\log2}\leqslant f(x)\leqslant\frac{\mathrm e^x}{\log x-\log2}+\mathrm e^x\frac{\mathrm e^{-cx}}{\log2}. $$ Finally, $$ f(x)=\frac{\mathrm e^x}{\log x}\,\left(1+O\left(\frac1{\log x}\right)\right). $$ Note: With some more care, one can replace the $O\left(\frac1{\log x}\right)$ error term by $O\left(x^{a-1/2}\right)$, for every positive $a$.