$sx^{-s-1}\,dx=d(1-x^{-s})\newcommand{\lint}{\operatorname{li}}$ (note the sign) doesn't help because $(1-x^{-s})\lint(x)\underset{x\to\infty}{\longrightarrow}\infty$ as well.
Rather, if we define $f(s)=\int_1^\infty sx^{-s-1}\lint(x)\,dx$ for $\Re s>1$, then $$f(s)-f(t)\underset{\big[\text{IBP}\big]}{=}\int_1^\infty\frac{x^{-s}-x^{-t}}{\log x}\,dx\underset{\big[x=e^y\big]}{=}\log\frac{t-1}{s-1}$$ using a Frullani integral, thus (as noted by @reuns) it suffices to show that $f(2)=0$.
Here is a possible way to do this. For $a>1$ we have
\begin{align*}
\lint(a)&=\lim_{\epsilon\to 0^+}\left(\int_{1+\epsilon}^a\frac{dt}{\log t}+\int_0^{1-\epsilon}\frac{dt}{\log t}\right)
\\&=\lim_{\epsilon\to 0^+}\left(\int_{1+\epsilon}^a\frac{dx}{\log x}-\int_{1/(1-\epsilon)}^\infty\frac{x^{-2}\,dx}{\log x}\right)
\\&=\lim_{\epsilon\to 0^+}\left(\int_{1+\epsilon}^a\frac{1-x^{-2}}{\log x}\,dx+\int_{1+\epsilon}^{1/(1-\epsilon)}\frac{x^{-2}\,dx}{\log x}-\int_a^\infty\frac{x^{-2}\,dx}{\log x}\right)
\\&=\int_1^a\frac{1-x^{-2}}{\log x}\,dx-\int_a^\infty\frac{x^{-2}\,dx}{\log x},
\end{align*}
so that, integrating by parts (again),
\begin{align*}
f(2)&=\lim_{a\to 1^+}\int_a^\infty 2x^{-3}\lint(x)\,dx
\\&=\lim_{a\to 1^+}\left(a^{-2}\lint(a)+\int_a^\infty\frac{x^{-2}\,dx}{\log x}\right)
\\&=\lim_{a\to 1^+}\left((a^{-2}-1)\lint(a)+\int_1^a\frac{1-x^{-2}}{\log x}\,dx\right).
\end{align*}
And both terms tend to zero (use $\lint(a)=\log(a-1)+O(1)$ as $a\to 1^+$).
Another solution is to substitute $x=e^t$ and use the known series
$$
\lint(e^t)=\gamma+\log t+\sum_{n=1}^\infty\frac{t^n}{n!\ n}\qquad(t>0)
$$
so that, for $\Re(s)>1$, termwise integration yields
\begin{align*}
&\color{white}{=}s\int_1^\infty\lint(x)x^{-s-1}\,dx
=s\int_0^\infty\lint(e^t)e^{-st}\,dt
\\&=s\int_0^\infty(\gamma+\log t)e^{-st}+\sum_{n=1}^\infty\frac{s}{n!\ n}\int_0^\infty t^n e^{-st}\,dt
\\&=\gamma+s\int_0^\infty e^{-st}\log t\,dt+\sum_{n=1}^\infty\frac1{ns^n}
\\&=-\log s-\log(1-1/s)=-\log(s-1)
\end{align*}
where we use
$$
s\int_0^\infty e^{-st}\log t\,dt=\int_0^\infty e^{-x}\log(x/s)\,dx=-\gamma-\log s.
$$
