It can be shown that
$$ \begin{align} \int_0^{\infty } \frac{e^{-\rho u-\frac{v}{\rho }}}{2\rho } \, d\rho=K_0(2\sqrt{uv}) \end{align} $$ for $u,v>0$. Now, I wish to have a similar integral expression for $I_0(\sqrt{u v})$ (and, or, $J_0(\sqrt{u v})$), however $I_0(x)$ (and therefore $J_0(x)$) cannot be expressed in terms of $K_0(x)$, so I cannot transform the previous integral to this end. The question is then, are there constants $a$ and $b$ (maybe complex) and a function $C(\rho)$ such that
$$ \begin{align} \int_0^{\infty } C(\rho)\:e^{a\rho u+b\frac{v}{\rho }} \, d\rho=I_0(2\sqrt{uv}) \end{align} $$
$ab=1$ since $\partial^2I_0(2\sqrt{uv})/\partial u\partial v=I_0(2\sqrt{uv})$, but finding $C(\rho)$ is not that evident.
Taking $\nu=0$, $z=2\sqrt{uv}$ and $t = u\rho$ in Schläfli's integral http://dlmf.nist.gov/10.9.E19 yields $$ J_0 (2\sqrt {uv} ) = \frac{1}{{\pi i}}\int_{ - \infty }^{(0 + )} {\frac{{e^{\rho u - \frac{v}{\rho }} }}{{2\rho }}d\rho } . $$ In this case you are forced to integrate on a loop contour in the complex plane. Similarly, taking $\nu=0$, $z=2\sqrt{uv}$ and $e^t = \sqrt {u/v} \rho$ in http://dlmf.nist.gov/10.32.E12 gives $$ I_0 (2\sqrt {uv} ) = \frac{1}{{\pi i}}\int_{ - \infty }^{(0 + )} {\frac{{e^{\rho u + \frac{v}{\rho }} }}{{2\rho }}d\rho } . $$ Let me add that your claim that $J_0$ cannot be expressed in terms of $K_0$ is not true, cf. http://dlmf.nist.gov/10.27.E9.