Hint:
After splitting up the interval of integration, changing the order of integration and exploiting symmetry we find
$$\begin{align}
I
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\ln{\left(\left|x-y\right|\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(\left|x-y\right|\right)}+\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}y\,\ln{\left(\left|x-y\right|\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}+\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}y\,\ln{\left(y-x\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}+\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}x\,\ln{\left(y-x\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}+\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}\\
&=2\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\ln{\left(x-y\right)}.\\
\end{align}$$