Note that the integral $F(s)$ diverges at infinity for $s\leqslant1$ and redefine $F(s)$ for every $s\gt1$ as
$$
F(s)=\int\limits_2^{+\infty}\frac{\text{Li}(x)}{x^{s+1}}\mathrm dx.
$$
An integration by parts yields
$$
F(s)=\frac1s\int\limits_2^{+\infty}\frac{\mathrm dx}{x^s\log x},
$$
and the change of variable $x^{s-1}=\mathrm e^t$ yields $sF(s)=G((s-1)\log2)$ with
$$
G(z)=\int\limits_z^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t.
$$
For every $z\leqslant1$,
$$
G(z)=-\log(z)-\int\limits_z^{1}(1-\mathrm e^{-t})\frac{\mathrm dt}t+\int\limits_1^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t.
$$
Thus $F(s)=-\log(s-1)+H(s-1)$ where $H(z)$ can be expanded as a series in $z^n$ and $z^n\log(z)$ for nonnegative $n$, and
$$
H(0)=-\int\limits_0^{1}(1-\mathrm e^{-t})\frac{\mathrm dt}t+\int\limits_1^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t=-\mathrm{Ein}(1)+\mathrm{E}_1(1)=-\gamma,
$$
where $\mathrm{Ein}$ and $\mathrm{E}_1$ are related to the exponential integral function. More generally,
$$
G(z)=-\log(z)-\mathrm{Ein}(z)+\mathrm{E}_1(1)=-\log(z)-\gamma+\mathrm{Ein}(1)-\mathrm{Ein}(z),
$$
with
$$
\mathrm{Ein}(z) = \sum\limits_{k=1}^\infty \frac{(-1)^{k+1}}{k\,k!}z^k.
$$
Finally, using the expansion $\dfrac1s=\sum\limits_{n=0}^{+\infty}(-1)^n(s-1)^n$, one gets
$$
F(s)=-\log(s-1)+\sum\limits_{n=1}^{+\infty}(-1)^{n+1}(s-1)^n\log(s-1)-\gamma+\sum\limits_{n=1}^{+\infty}c_n(s-1)^n,
$$
for some coefficients $(c_n)_{n\geqslant1}$. Due to the log terms, this is a slightly more complicated expansion than the one suggested in the question, in particular $s\mapsto F(s)+\log(s-1)$ is not analytic around $s=1$.
