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, $$S =\frac 12+\frac 12\sum_{k=1}^\infty\frac{(1-2u)^k}{k}\,{}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right). $$ In the final sum, we write $${}_2F_1\left(\begin{matrix}-k,k\\2\end{matrix};-\frac{u^2}{1-2u}\right)=\sum_{j=0}^{k}\frac{(-k)_j(k)_j}{j!(2)_j}\left(-\frac{u^2}{1-2u}\right)^j. $$ 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 $$S=\frac 12-\frac 12\,\log(2u)-\frac 14+\frac{\log 2}2=\frac14-\frac12\log(u). $$