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.