Note that the integral $F(s)$ diverges at infinity for $s\leqslant1$ and redefine $F(s)$ for every $s\gt1$ as $ F(s)=\int_2^{+\infty}\frac{\text{Li}(x)}{x^{s+1}}\mathrm dx. $ An integration by parts yields $ sF(s)=\int_2^{+\infty}\frac{\mathrm dx}{x^s\log x}, $ and the change of variable $x^{s-1}=\mathrm e^t$ yields $sF(s)=\mathrm{E}_1(u\log2),\qquad u=s-1,$ where the exponential integral function $\mathrm{E}_1$ is defined, for every complex $z$ not a nonpositive real number, by $ \mathrm{E}_1(z)=\int_z^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t. $ One knows that, for every such $z$, $ \mathrm{E}_1(z) = -\gamma-\log z-\sum\limits_{k=1}^\infty \frac{(-z)^{k}}{k\,k!}. $ On the other hand, $\frac1s=\frac1{1+u}=\sum_{n\geqslant0}(-1)^nu^n,$ hence $F(s)=\frac1s\mathrm{E}_1(u\log2)=\sum_{n\geqslant0}(-1)^nu^n\cdot\left(-\gamma-\log\log2-\log u-\sum\limits_{k=1}^\infty \frac{(-1)^{k}(\log2)^k}{k\,k!}u^k\right). $ One sees that $F(1+u)$ coincides with a series in $u^n$ and $u^n\log u$ for nonnegative $n$, and that $G(u)=F(1+u)+\log u$ is such that $G(0)=-\gamma-\log\log2.$ Finally, $ F(s)=-\gamma-\log\log2-\log(s-1)-\sum\limits_{n=1}^{+\infty}(-1)^{n}(s-1)^n\log(s-1)+\sum\limits_{n=1}^{+\infty}c_n(s-1)^n, $ for some coefficients $(c_n)_{n\geqslant1}$. Due to the logarithmic terms, this is a slightly more complicated expansion than the one suggested in the question, in particular $s\mapsto G(s-1)=F(s)+\log(s-1)$ is not analytic around $s=1$.