There is a closed form for $\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)} {{\it J}_{0}\left(\,\beta\,n\right)}}{{n}^{2}}}$? Using the method showed here proposed by Olivier Oloa with simplifications proposed by Anastasiya-Romanova, it is possible to prove that
$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,2\,n\right)}
{{\it J}_{0}\left(\,n\right)}}{{n}^{2}}}={\frac {5}{8}}+{\frac {1}{6}}\,{\pi }^{2}-4\,{\frac {
{\it EllipticE} \left( {\frac {1}{2}} \right) }{\pi }}-\frac{2}{{\pi }}\,\int _{0}^{1}\!{\frac {
v\arcsin \left( {\frac {1}{2}}\,v \right) }{\sqrt {1-{v}^{2}}}}{dv}
$$
but for the last integral it is  possible to have a closed form given by  Jack D'Aurizio (Please note that Jack is using the notation for elliptic functions in Mathematica and I am using the more standard notation used by Maple).  Then we have
$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,2\,n\right)}
{{\it J}_{0}\left(\,n\right)}}{{n}^{2}}}=-8\,{\frac {{\it EllipticE}
 \left( \frac {1}{2} \right) }{\pi }}+3\,{\frac {{\it EllipticK} \left( \frac {1}{2}
 \right) }{\pi }}+\frac {5}{8}+\frac {1}{6}\,{\pi }^{2}
$$
According with this result the answer to my question maybe yes. Do you agree?
 A: Using the method showed here proposed by Olivier Oloa with simplifications proposed by Anastasiya-Romanova and which is essentially the method that David H is showing, I am obtaining
$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)}
{{\it J}_{0}\left(\,\beta\,n\right)}}{{n}^{2}}}=\frac {{\beta}^{2}}{8}\,+\frac{1}{6}\,{
\pi }^{2}+\frac{1}{8}\,{\alpha}^{2}-2\,{\alpha}^{2}{\it EllipticF} \left( {
\frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\beta}^{-1}{
\pi }^{-1}+2\,\beta\,{\it EllipticF} \left( {\frac {\beta}{\alpha}},{
\frac {\alpha}{\beta}} \right) {\pi }^{-1}-2\,\beta\,{\it EllipticE}
 \left( {\frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\pi }
^{-1}-2\,\beta{\pi }^{-1}\,\int _{0}^{1}\!v\arcsin \left( {\frac {\beta\,v}{
\alpha}} \right) {\frac {1}{\sqrt {1-{v}^{2}}}}{dv}
$$ 
Then there is a closed form if the last integral has a closed form.  But for the last integral it is possible to have a closed form given by Daniel H (Please note that Daniel is using the notation for elliptic functions in Mathematica and I am using the more standard notation used by Maple). Then we have the following closed form:
$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)}
{{\it J}_{0}\left(\,\beta\,n\right)}}{{n}^{2}}}=\frac{1}{8}\,{\beta}^{2}+\frac{1}{6}\,{
\pi }^{2}+\frac{1}{8}\,{\alpha}^{2}-2\,{\alpha}^{2}{\it EllipticF} \left( {
\frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\beta}^{-1}{
\pi }^{-1}+2\,\beta\,{\it EllipticF} \left( {\frac {\beta}{\alpha}},{
\frac {\alpha}{\beta}} \right) {\pi }^{-1}-2\,\beta\,{\it EllipticE}
 \left( {\frac {\beta}{\alpha}},{\frac {\alpha}{\beta}} \right) {\pi }
^{-1}-2\,{\beta}^{2}{\it EllipticK} \left( {\frac {\beta}{\alpha}}
 \right) {\pi }^{-1}{\alpha}^{-1}+2\,\alpha\,{\it EllipticK} \left( {
\frac {\beta}{\alpha}} \right) {\pi }^{-1}-2\,\alpha\,{\it EllipticE}
 \left( {\frac {\beta}{\alpha}} \right) {\pi }^{-1}
$$
It is worthwhile to note  the following particular case:
$$\sum _{n=1}^{\infty }{\frac {{{\it J}_{0}\left(\,\alpha\,n\right)}
{{\it J}_{0}\left(\,\frac{1}{2}\,\alpha\,n\right)}}{{n}^{2}}}={\frac {5}{32}}\,{
\alpha}^{2}+\frac{1}{6}\,{\pi }^{2}-3\,{\frac {\alpha\,{\it EllipticF} \left( 
\frac{1}{2},2 \right) }{\pi }}-{\frac {\alpha\,{\it EllipticE} \left( \frac{1}{2},2
 \right) }{\pi }}-2\,{\frac {\alpha\,{\it EllipticE} \left( \frac{1}{2}
 \right) }{\pi }}+\frac{3}{2}\,{\frac {\alpha\,{\it EllipticK} \left( \frac{1}{2}
 \right) }{\pi }}
$$
A: (Too long for comment)
The Bessel function of the first kind of order zero, $J_{0}{\left(z\right)}$, has the following integral representation:
$$J_{0}{\left(z\right)}=\frac{2}{\pi}\int_{0}^{1}\mathrm{d}t\,\frac{\cos{\left(zt\right)}}{\sqrt{1-t^2}}.$$
Represent the Bessel functions in the series by the integrals,
$$J_{0}{\left(\alpha n\right)}=\frac{2}{\pi}\int_{0}^{1}\mathrm{d}x\,\frac{\cos{\left(\alpha nx\right)}}{\sqrt{1-x^2}},$$
$$J_{0}{\left(\beta n\right)}=\frac{2}{\pi}\int_{0}^{1}\mathrm{d}y\,\frac{\cos{\left(\beta ny\right)}}{\sqrt{1-y^2}}.$$
Assume that $0<\alpha,\beta$ and $\alpha+\beta<2\pi$. Then,
$$\sum_{n=1}^{\infty}\frac{\cos{\left[\left(\alpha x+\beta y\right)n\right]}}{n^2}=\frac{\pi^2}{6}-\frac{\pi \left(\alpha x+\beta y\right)}{2}+\frac{\left(\alpha x+\beta y\right)^2}{4}.$$
We can then write the series as a double integral:
$$\begin{align}
S{\left(\alpha,\beta\right)}
&=\sum_{n=1}^{\infty}\frac{J_{0}{\left(\alpha n\right)} J_{0}{\left(\beta n\right)}}{n^2}\\
&=\frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2} \int_{0}^{1}\mathrm{d}x\,\frac{\cos{\left(\alpha nx\right)}}{\sqrt{1-x^2}} \int_{0}^{1}\mathrm{d}y\,\frac{\cos{\left(\beta ny\right)}}{\sqrt{1-y^2}}\\
&=\frac{4}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\sum_{n=1}^{\infty}\frac{\cos{\left(\alpha nx\right)}\,\cos{\left(\beta ny\right)}}{n^2\,\sqrt{1-x^2}\,\sqrt{1-y^2}}\\
&=\frac{4}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left(\alpha nx\right)}\,\cos{\left(\beta ny\right)}}{n^2}\\
&=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}+\cos{\left[(\alpha x+\beta y)n\right]}}{n^2}\\
&=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\
&~~~~~ + \frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x+\beta y)n\right]}}{n^2}\\
&=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\
&~~~~~ + \frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\left[\frac{\pi^2}{6}-\frac{\pi \left(\alpha x+\beta y\right)}{2}+\frac{\left(\alpha x+\beta y\right)^2}{4}\right]\\
&=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\
&~~~~~ + \frac{2}{\pi^2}\left[\frac{\pi^4}{24}-\frac{\pi^2}{4}\left(\alpha+\beta\right)+\frac12\alpha\beta+\frac{\pi^2}{32}\left(\alpha^2+\beta^2\right)\right]\\
&=\frac{2}{\pi^2}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{1-x^2}\,\sqrt{1-y^2}}\sum_{n=1}^{\infty}\frac{\cos{\left[(\alpha x-\beta y)n\right]}}{n^2}\\
&~~~~~ + \left[\frac{\pi^2}{12}-\frac{1}{2}\left(\alpha+\beta\right)+\frac{\alpha\beta}{\pi^2}+\frac{1}{16}\left(\alpha^2+\beta^2\right)\right]\\
\end{align}$$
...
