Here's a hand-wavey argument using the generating function $e^{\frac{x}{2}(t-1/t)} = \sum_{m=-\infty}^\infty t^m J_m(z)$:
$\begin{eqnarray} e^{\frac{x}{2}(t-1/t)+\frac{x}{2}(u-1/u)} &=& e^{\frac{x}{2}(t-1/t)}e^{\frac{x}{2}(u-1/u)}\\\ &=& \left(\sum_{m=-\infty}^\infty t^m J_m(x)\right)\left(\sum_{n=-\infty}^\infty u^n J_n(x)\right)\\\ &=&\sum_{m=-\infty}^\infty\;\sum_{n=-\infty}^\infty t^{m}u^{n}J_m(x)J_n(x)\\\ &=&\sum_{m=-\infty}^\infty\;\sum_{k=-\infty}^\infty t^{m}u^{m+k}J_m(x)J_{m+k}(x)\\\ &=&\sum_{k=-\infty}^\infty u^k\sum_{m=-\infty}^\infty (tu)^{m}J_m(x)J_{m+k}(x)\;. \end{eqnarray}$
Now let $t=1/u$ and we get
$1 = \sum_{k=-\infty}^\infty u^k \left(\sum_{m=-\infty}^\infty J_m(x)J_{m+k}(x)\right)\;,$
which gives the desired result (since the parenthesized term is independent of $u$).
[Thanks to Joriki for cleaning up the LaTeX.]