It seems I figured it out. 8.111 in Lewin's book has the form $$\int\limits_0^x\frac{\ln{(1-y)}\ln{(1-cy)}}{y}\,dy=\mathrm{Li}_3\left(\frac{1-xc}{1-x}\right)+\mathrm{Li}_3\left(\frac{1}{c}\right)+\mathrm{Li}_3(1)-\mathrm{Li}_3(1-cx)-\mathrm{Li}_3(1-x)-\mathrm{Li}_3\left(\frac{1-xc}{c(1-x)}\right)+\ln{(1-cx)}\left[\mathrm{Li}_2\left(\frac{1}{c}\right)-\mathrm{Li}_2(x)\right]$$ $$+\ln{(1-x)}\left[\mathrm{Li}_2(1-cx)-\mathrm{Li}_2\left(\frac{1}{c}\right)+\zeta(2)\right]+\frac{1}{2}\ln{(c)}\ln^2{(1-x)}.$$ However there is a subtlety here. In l.h.s. of (1) we assume that for $y>1$, $\ln{(1-y)}=i\pi+\ln{(y-1)}$, that is that we are above the branch cut of the logarithmic function. Therefore its correct form is $$\ln{(1-y+i\epsilon)}=\ln{[1-(y-i\epsilon)]}=i\pi+\ln{(y-1)},$$ which suggests to use $x=2-i\epsilon$, rather than $x=2$ in 8.111 (along with $c=-1$). As a result we get (note that $\mathrm{Li}_3(1)=\zeta(3)$, $\mathrm{Li}_2(-1)=-\zeta(2)/2$) \begin{eqnarray} && J=\int\limits_0^{2-i\epsilon}\frac{\ln{(1-y)}\ln{(1+y)}}{y}\,dy=\zeta(3)+ \mathrm{Li}_3(-3)-\left[\mathrm{Li}_3(3-i\epsilon)+\mathrm{Li}_3(3+i\epsilon)\right]+\\ && \ln{3}\left[-\frac{1}{2}\,\zeta(2)-\mathrm{Li}_2(2-i\epsilon)\right]+i\pi\left[\mathrm{Li}_2(3-i\epsilon)-\frac{3}{2}\,\zeta(2)\right]. \end{eqnarray} Using the functional relation $$\mathrm{Li}_2(y)+\mathrm{Li}_2(1-y)=\zeta(2)-\ln{y}\ln{(1-y)},$$ we get $$\mathrm{Li}_2(2-i\epsilon)=-\mathrm{Li}_2(-1)+\zeta(2)-\ln{(-1+i\epsilon)}\ln{2}=\frac{3}{2}\,\zeta(2)-i\pi\ln{2},$$ and $$\mathrm{Li}_2(3-i\epsilon)=-\mathrm{Li}_2(-2)+\zeta(2)-\ln{(-2+i\epsilon)}\ln{3}=-\mathrm{Li}_2(-2)+\zeta(2)-\ln{2}\ln{3}-i\pi\ln{3}.$$ So that the integral takes the form \begin{equation} J=\zeta(3)+\mathrm{Li}_3(-3)-\left[\mathrm{Li}_3(3-i\epsilon)+\mathrm{Li}_3 (3+i\epsilon)\right]+4\zeta(2)\ln{3}-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2)\right] \end{equation} Now we use the inversion equation $$\mathrm{Li}_3(-y)-\mathrm{Li}_3\left(-\frac{1}{y}\right)=-\frac{1}{6}\, \ln^3{y}+2\mathrm{Li}_2(-1)\ln{y},$$ and its complex variant $$\mathrm{Li}_3\left(\frac{1}{z}\right)=\mathrm{Li}_3(z)+\frac{1}{6}\, \ln^3{(-z)}+\zeta(2)\ln{(-z)},\;\;|z|<1,$$ to get \begin{equation} \mathrm{Li}_3(-3)=\mathrm{Li}_3\left(-\frac{1}{3}\right)-\frac{1}{6}\,\ln^3{3}-\zeta(2)\ln{3}, \tag{2} \end{equation} and $$\mathrm{Li}_3(3\pm i\epsilon)=\mathrm{Li}_3\left(\frac{1}{3}\right)- \frac{1}{6}\,\ln^3{3}+2\zeta(2)\ln{3}\pm\frac{i\pi}{2}\,\ln^2{3},$$ so that $$\mathrm{Li}_3(3+i\epsilon)+\mathrm{Li}_3(3-i\epsilon)=2\mathrm{Li}_3\left(\frac{1}{3}\right)-\frac{1}{3}\,\ln^3{3}+4\zeta(2)\ln{3},$$ and \begin{equation} J=\zeta(3)+\mathrm{Li}_3\left(-\frac{1}{3}\right)-2\mathrm{Li}_3\left(\frac{1}{3}\right)+\frac{1}{6}\,\ln^3{3}-\zeta(2)\ln{3}-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2)\right]. \tag{3} \end{equation} Functional equation connecting five trilogarithms $$\mathrm{Li}_3\left(\frac{1-x}{1+x}\right)-\mathrm{Li}_3\left(-\frac{1-x} {1+x}\right)=2\mathrm{Li}_3(1-x)+2\mathrm{Li}_3\left(\frac{1}{1+x}\right)- \frac{1}{2}\mathrm{Li}_3(1-x^2)-\frac{7}{4}\zeta(3)+\zeta(2)\ln{(1+x)}- \frac{1}{3}\,\ln^3{(1+x)},$$ gives for $x=2$ $$\mathrm{Li}_3\left(-\frac{1}{3}\right)-3\mathrm{Li}_3\left(\frac{1}{3} \right)=-\frac{13}{4}\,\zeta(3)-\frac{1}{2}\,\mathrm{Li}_3(-3)+\zeta(2)\ln{3}-\frac{1}{3}\,\ln^3{3},$$ where we have used $\mathrm{Li}_3(-1)=-\frac{3}{4}\,\zeta(3)$. In the r.h.s. of this equation , $\mathrm{Li}_3(-3)$ can be expressed through (2) and we get after some simple algebra $$\mathrm{Li}_3\left(-\frac{1}{3}\right)-2\mathrm{Li}_3\left(\frac{1}{3} \right)=-\frac{13}{6}\,\zeta(3)-\frac{1}{6}\,\ln^3{3}+\zeta(2)\ln{3}.$$ Therefore (3) takes the form $$J=-\frac{7}{6}\,\zeta(3)-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2)\right].$$

If we naively assume that in $\mathrm{Li}_3(3-i\epsilon)+\mathrm{Li}_3(3+i\epsilon)$ the first term is also above the cut, we erroneously get an extra imaginary part $i\pi\zeta(2)\ln^2{3}$.

Note that the calculated integral can be used to evaluate another integral which surfaced in this MO question: An interesting integral expression for $\pi^n$? $$\int\limits_0^1\frac{\ln{x}\ln{(2+x)}}{1+x}dx=$$ $$\int\limits_1^2\frac{\ln{(y-1)}\ln{(1+y)}}{y}dy=\int\limits_1^2\frac{\ln{(1-y)}\ln{(1+y)}}{y}dy-i\pi\int\limits_1^2\frac{\ln{(1+y)}}{y}dy,$$ and using $$\int\limits_0^1\frac{\ln{(1+y)}}{y}dy=\frac{1}{2}\zeta(2),\;\; \int\limits_0^2\frac{\ln{(1+y)}}{y}dy=-\mathrm{Li}_2(-2),$$ and $$\int\limits_0^1\frac{\ln{(1-y)}\ln{(1+y)}}{y}dy=-\frac{5}{8}\,\zeta(3),\;\;\;\;\; \int\limits_0^2\frac{\ln{(1-y)}\ln{(1+y)}}{y}dy=-\frac{7}{6}\,\zeta(3)-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2) \right],$$ we get $$\int\limits_0^1\frac{\ln{x}\ln{(2+x)}}{1+x}dx=\left[\frac{5}{8}-\frac{7}{6}\right]\zeta(3)=-\frac{13}{24}\,\zeta(3).$$

It seems I figured it out. 8.111 in Lewin's book has the form $$\int\limits_0^x\frac{\ln{(1-y)}\ln{(1-cy)}}{y}\,dy=\mathrm{Li}_3\left(\frac{1-xc}{1-x}\right)+\mathrm{Li}_3\left(\frac{1}{c}\right)+\mathrm{Li}_3(1)-\mathrm{Li}_3(1-cx)-\mathrm{Li}_3(1-x)-\mathrm{Li}_3\left(\frac{1-xc}{c(1-x)}\right)+\ln{(1-cx)}\left[\mathrm{Li}_2\left(\frac{1}{c}\right)-\mathrm{Li}_2(x)\right]$$ $$+\ln{(1-x)}\left[\mathrm{Li}_2(1-cx)-\mathrm{Li}_2\left(\frac{1}{c}\right)+\zeta(2)\right]+\frac{1}{2}\ln{(c)}\ln^2{(1-x)}.$$ However there is a subtlety here. In l.h.s. of (1) we assume that for $y>1$, $\ln{(1-y)}=i\pi+\ln{(y-1)}$, that is that we are above the branch cut of the logarithmic function. Therefore its correct form is $$\ln{(1-y+i\epsilon)}=\ln{[1-(y-i\epsilon)]}=i\pi+\ln{(y-1)},$$ which suggests to use $x=2-i\epsilon$, rather than $x=2$ in 8.111 (along with $c=-1$). As a result we get (note that $\mathrm{Li}_3(1)=\zeta(3)$, $\mathrm{Li}_2(-1)=-\zeta(2)/2$) \begin{eqnarray} && J=\int\limits_0^{2-i\epsilon}\frac{\ln{(1-y)}\ln{(1+y)}}{y}\,dy=\zeta(3)+ \mathrm{Li}_3(-3)-\left[\mathrm{Li}_3(3-i\epsilon)+\mathrm{Li}_3(3+i\epsilon)\right]+\\ && \ln{3}\left[-\frac{1}{2}\,\zeta(2)-\mathrm{Li}_2(2-i\epsilon)\right]+i\pi\left[\mathrm{Li}_2(3-i\epsilon)-\frac{3}{2}\,\zeta(2)\right]. \end{eqnarray} Using the functional relation $$\mathrm{Li}_2(y)+\mathrm{Li}_2(1-y)=\zeta(2)-\ln{y}\ln{(1-y)},$$ we get $$\mathrm{Li}_2(2-i\epsilon)=-\mathrm{Li}_2(-1)+\zeta(2)-\ln{(-1+i\epsilon)}\ln{2}=\frac{3}{2}\,\zeta(2)-i\pi\ln{2},$$ and $$\mathrm{Li}_2(3-i\epsilon)=-\mathrm{Li}_2(-2)+\zeta(2)-\ln{(-2+i\epsilon)}\ln{3}=-\mathrm{Li}_2(-2)+\zeta(2)-\ln{2}\ln{3}-i\pi\ln{3}.$$ So the integral takes the form \begin{equation} J=\zeta(3)+\mathrm{Li}_3(-3)-\left[\mathrm{Li}_3(3-i\epsilon)+\mathrm{Li}_3 (3+i\epsilon)\right]+4\zeta(2)\ln{3}-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2)\right] \end{equation} Now we use the inversion equation $$\mathrm{Li}_3(-y)-\mathrm{Li}_3\left(-\frac{1}{y}\right)=-\frac{1}{6}\, \ln^3{y}+2\mathrm{Li}_2(-1)\ln{y},$$ and its complex variant $$\mathrm{Li}_3\left(\frac{1}{z}\right)=\mathrm{Li}_3(z)+\frac{1}{6}\, \ln^3{(-z)}+\zeta(2)\ln{(-z)},\;\;|z|<1,$$ to get \begin{equation} \mathrm{Li}_3(-3)=\mathrm{Li}_3\left(-\frac{1}{3}\right)-\frac{1}{6}\,\ln^3{3}-\zeta(2)\ln{3}, \tag{2} \end{equation} and $$\mathrm{Li}_3(3\pm i\epsilon)=\mathrm{Li}_3\left(\frac{1}{3}\right)- \frac{1}{6}\,\ln^3{3}+2\zeta(2)\ln{3}\pm\frac{i\pi}{2}\,\ln^2{3},$$ so that $$\mathrm{Li}_3(3+i\epsilon)+\mathrm{Li}_3(3-i\epsilon)=2\mathrm{Li}_3\left(\frac{1}{3}\right)-\frac{1}{3}\,\ln^3{3}+4\zeta(2)\ln{3},$$ and \begin{equation} J=\zeta(3)+\mathrm{Li}_3\left(-\frac{1}{3}\right)-2\mathrm{Li}_3\left(\frac{1}{3}\right)+\frac{1}{6}\,\ln^3{3}-\zeta(2)\ln{3}-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2)\right]. \tag{3} \end{equation} Functional equation connecting five trilogarithms $$\mathrm{Li}_3\left(\frac{1-x}{1+x}\right)-\mathrm{Li}_3\left(-\frac{1-x} {1+x}\right)=2\mathrm{Li}_3(1-x)+2\mathrm{Li}_3\left(\frac{1}{1+x}\right)- \frac{1}{2}\mathrm{Li}_3(1-x^2)-\frac{7}{4}\zeta(3)+\zeta(2)\ln{(1+x)}- \frac{1}{3}\,\ln^3{(1+x)},$$ gives for $x=2$ $$\mathrm{Li}_3\left(-\frac{1}{3}\right)-3\mathrm{Li}_3\left(\frac{1}{3} \right)=-\frac{13}{4}\,\zeta(3)-\frac{1}{2}\,\mathrm{Li}_3(-3)+\zeta(2)\ln{3}-\frac{1}{3}\,\ln^3{3},$$ where we have used $\mathrm{Li}_3(-1)=-\frac{3}{4}\,\zeta(3)$. In the r.h.s. of this equation , $\mathrm{Li}_3(-3)$ can be expressed through (2) and we get after some simple algebra $$\mathrm{Li}_3\left(-\frac{1}{3}\right)-2\mathrm{Li}_3\left(\frac{1}{3} \right)=-\frac{13}{6}\,\zeta(3)-\frac{1}{6}\,\ln^3{3}+\zeta(2)\ln{3}.$$ Therefore (3) takes the form $$J=-\frac{7}{6}\,\zeta(3)-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2)\right].$$

If we naively assume that in $\mathrm{Li}_3(3-i\epsilon)+\mathrm{Li}_3(3+i\epsilon)$ the first term is also above the cut, we erroneously get an extra imaginary part $i\pi\ln^2{3}$.

Note that the calculated integral can be used to evaluate another integral which surfaced in this MO question: An interesting integral expression for $\pi^n$? $$\int\limits_0^1\frac{\ln{x}\ln{(2+x)}}{1+x}dx=$$ $$\int\limits_1^2\frac{\ln{(y-1)}\ln{(1+y)}}{y}dy=\int\limits_1^2\frac{\ln{(1-y)}\ln{(1+y)}}{y}dy-i\pi\int\limits_1^2\frac{\ln{(1+y)}}{y}dy,$$ and using $$\int\limits_0^1\frac{\ln{(1+y)}}{y}dy=\frac{1}{2}\zeta(2),\;\; \int\limits_0^2\frac{\ln{(1+y)}}{y}dy=-\mathrm{Li}_2(-2),$$ and $$\int\limits_0^1\frac{\ln{(1-y)}\ln{(1+y)}}{y}dy=-\frac{5}{8}\,\zeta(3),\;\;\;\;\; \int\limits_0^2\frac{\ln{(1-y)}\ln{(1+y)}}{y}dy=-\frac{7}{6}\,\zeta(3)-i\pi\left[\mathrm{Li}_2(-2)+\frac{1}{2}\,\zeta(2) \right],$$ we get $$\int\limits_0^1\frac{\ln{x}\ln{(2+x)}}{1+x}dx=\left[\frac{5}{8}-\frac{7}{6}\right]\zeta(3)=-\frac{13}{24}\,\zeta(3).$$