The identity can also be shown without divergences using generating functions, as suggested in the comment of @TimothyBudd above.
Using the generating functions \begin{align} \tag{1a}\label{eq:1a} f(u,t)&=\frac{t-t^{-1}}{4u}\left( 1-\frac{1}{1-t}\sqrt{(1-t)^2+4 u t}\right) -\frac 1 2\\ \tag{1b}\label{eq:1b} &=\sum_{m=1}^\infty {}_{2}F_1(-m,m;2;u)\,t^m \end{align} as well as \begin{align} \tag{2a}\label{eq:2a} g(u,t)&=\int_0^{t} \mathrm d t'\frac{f(u,t')}{t'} \\ \tag{2a}\label{eq:2b} &=\frac{t-1}{t+1}\left(f(u,t)+\frac 1 2\right) +\log\left(2 \,\frac{f(u,t)+\frac 1 2}{t+1}\right) + \frac 1 2\\ \tag{2c}\label{eq:2c} &=\sum_{m=1}^\infty \frac{\,{}_{2}F_1(-m,m;2;u)}{m}\, t^m \,, \end{align} we can apply the Hadamard product formula at $t=1$ to get \begin{align} \tag{3a}\label{eq:3a} L(u)&= \sum_{m=1}^\infty \frac {\,{}_{2}F_1(-m,m;2;u)^2}{m} \\ \tag{3b}\label{eq:3b} &= \frac{1}{2\pi i} \oint_{|z|=1} \mathrm d z \,z^{-1}f(u,z^{\pm 1})\,g(u,z^{\mp 1}) \,. \end{align} The analytic evaluation of this unit circle contour integral is left as an exercise. Numerically it gives the desired result.