Asymptotic expansion of $Li^{-1}$ and zeros of $F(s)$ and $G(s)$

First of all $$\mathrm{li}(z)=\mathrm{Ei}(\ln(z))$$ and the problem would be to find the inverse of the exponential integral function.

For that, you could find a very interesting approach in the paper entitled

"On the function inverse to the exponential integral function"

published by P. Pecina in

Astronomical Institutes of Czechoslovakia, Bulletin (ISSN 0004-6248), vol. 37, Jan. 1986, pp 8-12

Have a look here and you can access the pdf of the full paper which even contains the Fortran code for high accuracy.

On the other hand, if I properly remember,

$$Li^{-1}(n) \sim n \left(\log (n)+\log (\log (n))-1+\frac{\log (\log (n))-2}{\log (n)}+\cdots\right)$$

Edit

After @reuns's comment, I asked a former colleague of mine; he gave me the next term inside brackets $$-\frac{\log ^2(\log (n))-6 \log (\log (n))+11}{2 \log ^2(n)}$$ This was published by Cesaro (have a look here) in 1894.

So, in a more compact form

$$Li^{-1}(n) \sim n \left(L_1+L_2-1+\frac{L_2-2}{L_1}-\frac{L_2^2-6 L_2+11}{2 L_1^2}+\cdots\right)$$ where $L_1=\log(n)$ and $L_2=\log(L_1)$. The part in brackets "looks" like the expansion of Lambert function for infinitely large values of the argument.