Here I proposed a generalized formula for the polylogarithm. However, because of a slight mistake towards the end, visible prior to the edit, I was unaware that it yields just a result of an integral substitution. So I thought about the problem further.
Here I asked for verification of the derivation. Notice that in line $$ \frac{d^{c-1}}{dx^{c-1}}\frac{x^{c-1}}{1-x}Li_c(z) = \sum_{j=0}^{\infty}\sum_{n=j}^{\infty} {n \brace j}z^{j}\frac{x^{n}}{n!}(z j! \int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^{j+1}}dt+j(j-1)!\int_{0}^{\infty}\frac{t^{c+n-1}}{(e^t-z)^j}dt)$$ there have been Stirling numbers used. Thus in the next line the exponential generating function for them have been substituted. What if I modify the Stirling numbers?
In the answer to this post I gave a formula for modified Stirling numbers of the second kind. If one plugs it in, then one obtains instead $$ \sum_{n=0}^{\infty}\frac{d^{s-1}}{dx^{s-1}}x^{n+s-1}(z^q\frac{d}{dz})^nLi_{s+n}(z)=\int_0^{\infty} \frac{t^{s-1}dt}{\frac{1}{zf(txz^{q-1})}e^t-1} $$
provided that $f(0)=1$ and $f^q=f'$.
$q=2$ gives a $t$ next to the exponential in the denominator. This could be potentially helpful but I do not know how to exploit it if it is. Are there any values of the Polylogarithm at specific constants one could deduce thanks to this formula? $q$ can be any real number. Solutions to the differential equation give just the reciprocals of the $\frac{1}{q-1}$ powers.