Here is a generalization:
\begin{gather*}
\int_0^1\frac{\operatorname{Li}_{2a}(x)}{1+x}\mathrm{d}x=\int_0^1\frac{\operatorname{Li}_{2a}(x)}{x}\mathrm{d}x-\int_0^1\frac{\operatorname{Li}_{2a}(x)}{x(1+x)}\mathrm{d}x\\
=\operatorname{Li}_{2a+1}(x)|_0^1-\int_0^1\frac{1}{1+x}\left(\frac{-1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1-xy}\mathrm{d}y\right)\mathrm{d}x\\
=\operatorname{Li}_{2a+1}(1)+\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\int_0^1\frac{\mathrm{d}x}{(1+x)(1-yx)}\right)\mathrm{d}y\\
=\zeta(2a+1)+\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\frac{\ln(2)-\ln(1-y)}{1+y}\right)\mathrm{d}y\\
=\zeta(2a+1)+\frac{\ln(2)}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1+y}\mathrm{d}y\\
-\frac{1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)\ln(1-y)}{1+y}\mathrm{d}y.
\end{gather*}
Substitute
$$\int_0^1\frac{\ln^a(x)}{1+x}\mathrm{d}x=(-1)^aa!\,(1-2^{-a})\zeta(a+1)$$
and
\begin{gather*}
\int_0^1\frac{\ln^{2q-1}(x)\ln(1-x)}{1+x}\mathrm{d}x=(2^{1-2q}-2)(2q-1)!\ln(2)\zeta(2q)\\
+[q-2^{-1-2q}(1+2q-2^{1+2q})](2q-1)!\,\zeta(2q+1)\\
-(2q-1)!\sum_{k=1}^{q-1}(1-2^{2k-2q})\zeta(2k)\zeta(2q-2k+1),
\end{gather*}
we get
\begin{gather*}
\int_0^1\frac{\operatorname{Li}_{2a}(x)}{1+x}\mathrm{d}x=[1-a+2^{-1-2a}(2a+1-2^{1+2a})]\zeta(2a+1)\nonumber\\
+\ln(2)\zeta(2a)+\sum_{k=1}^{a-1}(1-2^{2k-2a})\zeta(2k)\zeta(2a-2k+1).
\end{gather*}
Examples
\begin{gather}
\int_0^1\frac{\operatorname{Li}_{2}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(2)-\frac58\zeta(3);\\
\int_0^1\frac{\operatorname{Li}_{4}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(4)+\frac34\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)\label{int_0^1 Li_4(x)/(1+x)};\\
\int_0^1\frac{\operatorname{Li}_{6}(x)}{1+x}\mathrm{d}x=\ln(2)\zeta(6)+\frac34\zeta(3)\zeta(4)+\frac{15}{16}\zeta(2)\zeta(5)-\frac{377}{128}\zeta(7).
\end{gather}
About your main integral:
$$\int_0^1\frac{\operatorname{Li}_{2}(-x)\operatorname{Li}_{2}(x)}{x}\mathrm{d}x=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\operatorname{Li}_{2}(x)\mathrm{d}x$$
$$=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)$$
$$=-\frac34\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$$
and this sum is well known.
