As stated in the comments, by following the Feynman's approach to integrals we have:
$$\begin{eqnarray*} I'(s) &=& \int_{0}^{1}\left(1+\frac{\log x}{1-x}-\frac{\log x}{2}\right)\frac{x^s}{1-x}dx\\ &=& \frac{1}{2}\zeta(2,s+1)+\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{1-x}dx\\&=&\frac{1}{2}\zeta(2,s+1)-\frac{1}{2s+2}\phantom{}_3 F_2\left(1,1,2;3,s+2;1\right),\end{eqnarray*}$$
but probably there is a nicer form for the second term.
As a matter of fact, by treating the second term this way:
$$\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{1-x}dx=\lim_{a\to 0^+}\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{(1-x)^{1-a}}$$
we get:
$$ I'(s)=\frac{1}{2}\psi'(s+1)+s\,\psi'(s)+\psi(s)-\psi(s+1)-1\tag{1}$$
where $\psi$ is the digamma function $\psi(s)=\frac{\Gamma'(s)}{\Gamma(s)}$. By integrating $(1)$ we have:
$$ I(s) =
I(1)+\frac{3\gamma+1}{2}-s+\frac{\psi(s+1)}{2}-\log\Gamma(s+1)+s\,\psi(s).\tag{2}$$
All we need to find a closed expression for $I(s)$ is now to evaluate $I(1)$ or $I(0)$, since by $(2)$:
$$ I(0)=I(1)+\gamma-\frac{1}{2}$$
holds. In virtue of $(2)$ we also have:
$$\lim_{s\to +\infty} I(s) = I(1) + \frac{3\gamma-\log(2\pi)}{2},$$
but since
$$ f(x)=\left(\frac{1}{\log x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{x}{1-x}$$
is a continuous, positive, increasing and bounded function ($f(x)\leq\frac{1}{12}$) on $(0,1)$, as long as $s\to+\infty$ we have $I(s)\leq\frac{1}{12s}$, so $\lim_{s\to +\infty}I(s)=0$. Hence we have:
$$ I(1) = \frac{\log(2\pi)-3\gamma}{2},\qquad I(0) = \frac{\log(2\pi)-\gamma-1}{2} \tag{3}$$
and:
$$ I(s) =
\frac{1+\log(2\pi)}{2}-s+\frac{\psi(s+1)}{2}-\log\Gamma(s+1)+s\,\psi(s).\tag{4}$$
As a bonus, by expanding $(4)$ in a neighbourhood of $+\infty$ and exploiting convexity, we can also show that: $$\frac{1}{12s+8}\leq I(s)\leq\frac{1}{12s+6}$$ holds for any $s\in\mathbb{R}^+$.