The identity can be derived by differentiating both sides with respect to $u$: The rhs \begin{align}\tag{1}\label{eq:1} R(u)=\frac 1 4-\frac 1 2 \log u \end{align} simply gives \begin{align}\tag{2}\label{eq:2} R'(u)=-\frac{1}{2u}. \end{align} The lhs $L(u)$ is first written as hypergeometric double series, \begin{align}\tag{3}\label{eq:3} L(u) = \sum_{m=1}^\infty \frac 1 m \sum_{k=0}^\infty \sum_{l=0}^\infty \frac{ (-m)_k (m)_k \,(-m)_l (m)_l}{ (2)_k \, (2)_l} \frac{u^{k+l}}{k! \, l!} , \end{align} such that \begin{align}\tag{4}\label{eq:4} L'(u) = \sum_{m=1}^\infty \frac 1 m \sum_{k=0}^\infty \sum_{l=0}^\infty (k+l) \frac{(-m)_k (m)_k (-m)_l (m)_l}{(2)_k \,(2)_l} \frac{u^{k+l-1}}{k! \, l!}. \end{align} At this point we can reorder the sums and first evaluate the sum over $m$, \begin{align}\tag{5}\label{eq:5} L'(u) &= \sum_{k=0}^\infty \sum_{l=0}^\infty \sum_{m=1}^\infty \frac {k+l} m \frac{(-m)_k (m)_k (-m)_l (m)_l}{(2)_k \,(2)_l} \frac{u^{k+l-1}}{k! \, l!}\\ \tag{6}\label{eq:6} &= -\sum_{k=0}^\infty \sum_{l=0}^\infty \frac{(k-1) (l-1) (-1)_k (-1)_l}{2 (2)_k \,(2)_l} u^{k+l-1}. \end{align} We are lucky now, as the only term contributing to this double sum is the term with $k=l=0$, such that \begin{align}\tag{7}\label{eq:7} L'(u) &= -\frac{1}{2u}. \end{align} Note that this simple result was not obtained without the initial differentiation, as then the $k{=}l{=}0$ term diverges. The remaining step is the determination of the integration constant, which can most easily evaluated at $u=1$, where the sum trivially gives $L(1)=R(1)=1/4$. Note that here only the first term $m=1$ contributes.