I think the proof below should work. Although it looks more complicated than the one of Fred Hucht, it doesn't involve divergent series.

I write
 \begin{equation}\tag{1}S(t)=\sum_{m=1}^\infty\frac {t^m}m\, {}_2F_1\left(\begin{matrix}-m,m\\2\end{matrix};u\right)^2,
 \end{equation}
 where $t^m$ is inserted to improve convergence.
We are interested in the value $S=S(1)$.

The key fact that I use is Bateman's product formula 
\begin{multline*}{}_2F_1\left(\begin{matrix}-m,a+m\\b\end{matrix};u\right)\,{}_2F_1\left(\begin{matrix}-m,a+m\\b\end{matrix};v\right)\\
=(-1)^m\frac{(1+a-b)_m}{(b)_m}\sum_{k=0}^m
\frac{(-m)_k(a+m)_k}{k!(1+a-b)_k}(1-u-v)^k\,{}_2F_1\left(\begin{matrix}-k,a+k\\b\end{matrix};-\frac{uv}{1-u-v}\right).
 \end{multline*}
In the case of interest, $a=0$ and $b=2$, we need to interpret
$$\frac{(1+a-b)_m}{(1+a-b)_k}=(1+a-b+k)_{m-k}=(k-1)_{m-k}, $$
which vanishes if $m\geq 2$ and $0\leq k\leq 1$. That is, for $m\geq 2$, 
\begin{multline}\label{b}{}_2F_1\left(\begin{matrix}-m,m\\2\end{matrix};u\right){}_2F_1\left(\begin{matrix}-m,m\\2\end{matrix};v\right)\\
=\frac{(-1)^m}{(2)_m}\sum_{k=2}^m
\frac{(-m)_k(m)_k(k-1)_{m-k}}{k!}(1-u-v)^k\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{uv}{1-u-v}\right).
 \end{multline}
 
 In $S(t)$,
 we isolate the term with $m=1$. In the remaining terms, 
 insert (1), change the order of summation and replace $m$ by $m+k$. After simplification, this gives 
\begin{multline*}S(t)=t\left(1-\frac u2\right)^2
+
\sum_{k=2}^\infty\frac{(2k-1)!}{k!(k+1)!}\,t^k (1-2u)^k\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right)\\
\times \sum_{m=0}^\infty\frac{(2k)_m(k-1)_m}{m!(k+2)_m}\,(-t)^m.\end{multline*}
Note that if we **formally** put $t=1$ and use Kummer's summation formula
$${}_2F_1\left(\begin{matrix}a,b\\1+a-b\end{matrix};-1\right)=\frac{\Gamma(1+a-b)\Gamma(1+a/2)}{\Gamma(1+a)\Gamma(1+a/2-b)}, $$
the sum in $m$ evaluates to $(k+1)!k!/(2k)!$. However, for $t=1$ this sum is divergent. To remedy this, we first apply the linear transformation
$${}_2F_1\left(\begin{matrix}2k,k-1\\k+2\end{matrix};-t\right)=(1+t)^{3-2k}\,{}_2F_1\left(\begin{matrix}2-k,3\\k+2\end{matrix};-t\right).
 $$
 We can then apply the terminating case of Kummer's formula, which gives the same result as was obtained formally:
$$\lim_{t\rightarrow 1}{}_2F_1\left(\begin{matrix}2k,k-1\\k+2\end{matrix};-t\right)=2^{3-2k}\,{}_2F_1\left(\begin{matrix}2-k,3\\k+2\end{matrix};-1\right)
=2^{3-2k}\frac{(4)_{k-2}}{(5/2)_{k-2}}=\frac{(k+1)!k!}{(2k)!}.
 $$
Assuming that we can interchange limit and summation (something needs to be checked here), it follows that
$$S=\left(1-\frac u2\right)^2+
\frac 12\sum_{k=2}^\infty\frac{(1-2u)^k}{k}\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right). $$
If we start the summation  at $k=1$, we add the term $(1-u/2)^2-1/2$. Thus,
$$S
=\frac 12+\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right). $$
We write the final ${}_2F_1$ as a sum over $j$.
We split of the term corresponding to $j=0$, which gives the contribution
$$\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}=-\frac 12\,\log(2u).$$
In the remaining terms, we replace $k$ by $k+j$ and change the order of summation. This gives
\begin{multline*}\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}\sum_{j=1}^{k}\frac{(-k)_j(k)_j}{j!(2)_j}\left(-\frac{u^2}{1-2u}\right)^j \\
=\frac 12\sum_{j=1}^\infty\frac{(2j-1)!}{j!(j+1)!}\,u^{2j}\sum_{k=0}^\infty\frac{(2j)_k}{k!}(1-2u)^k
=\frac 12\sum_{j=1}^\infty\frac{(2j-1)!}{j!(j+1)!}\frac 1{4^j}
=-\frac 14+\frac{\log 2}2,
\end{multline*}
 where we used the non-terminating binomial theorem to compute the sum over $k$.
 The final step can be obtained from the  Taylor expansion
 $$\sum_{j=1}^\infty\frac{(2j-1)!}{j!(j+1)!}\,x^j=\log(2)-\log(1+\sqrt{1-4x})+\frac{\sqrt{1-4x}+2x-1}{4x}.$$
 Collecting the different contributions finally gives
 $$S=\frac 12-\frac 12\,\log(2u)-\frac 14+\frac{\log 2}2=\frac14-\frac12\log(u). $$