The limit (as $s\downarrow1/2$) will not change if the double sum is replaced by the corresponding double integral, for which we find this, as desired: 

[![enter image description here][1]][1]

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Alternatively, the limit (as $s\downarrow1/2$) will not change if the double sum is replaced by the double integral 
$$I(s):=(2s-1)
\iint\limits_{x,y>0,\,x+y>1}\frac{dx\,dy}{x^s y^s(x+y)} \\
=(2s-1)\int_1^\infty \frac{du}u\,\int_0^u \frac{dx}{x^s (u-x)^s}
\\
=(2s-1)\int_1^\infty \frac{du}{u^{2s}}\,J(s)
=J(s),$$
where
$$J(s):=\int_0^1 \frac{dt}{t^s (1-t)^s}\to J(1/2)=\pi.$$
So, $I(s)\to\pi$, as desired. 


  [1]: https://i.sstatic.net/l8QbX.png