The following integral $$ \int_0^{+\infty} K_0(\alpha\sqrt{x^2+z^2})\, dx, \alpha>0, $$ can be computed according to Gradshteyn-Ryzhik 6.596 (3) taking $\nu=0$ and $\mu=-1/2$. Its value is $\frac{\pi}{2\sqrt{\alpha}} e^{-\alpha|z|}$.
My question is: what about the integral $$ \int_0^{+\infty} \cosh(\beta x) K_0(\alpha\sqrt{x^2+z^2})dx, \alpha>0, \beta \geq 0? $$
Assume that $\alpha>\beta$ and $|\arg z|<\frac{\pi}{4}$. Then using $(10.32.10)$ and interchanging the order of integrations, \begin{align*} & \int_0^{ + \infty } {\cosh (\beta x)K_0 (\alpha \sqrt {x^2 + z^2 } )\,\text{d}x} \\ & = \frac{1}{2}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{4t}}} \right)\int_0^{ + \infty } {\cosh (\beta x)\exp \left( { - \alpha ^2 \frac{{x^2 }}{{4t}}} \right)\text{d}x} \frac{{\text{d}t}}{t}} \\ & = \frac{{\sqrt \pi }}{{2\alpha }}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{4t}} + \left( {\frac{\beta }{\alpha }} \right)^2 t} \right)\frac{{\text{d}t}}{{t^{1/2} }}} \\ & = \frac{1}{2}\sqrt {\frac{\pi }{{\alpha ^2 - \beta ^2 }}} \int_0^{ + \infty } {\exp \left( { - s - \frac{{(\alpha ^2 - \beta ^2 )z^2 }}{{4s}}} \right)\frac{{\text{d}s}}{{s^{1/2} }}} \\ & = \sqrt {\frac{{\pi z}}{{2\sqrt {\alpha ^2 - \beta ^2 } }}} K_{ - 1/2} (\sqrt {\alpha ^2 - \beta ^2 } z) \\ & = \frac{\pi }{{2\sqrt {\alpha ^2 - \beta ^2 } }}\text{e}^{ - \sqrt {\alpha ^2 - \beta ^2 }z } . \end{align*} In the last step, we employed $(10.39.2)$. Note that your simplification in terms of an exponential when $\beta=0$ is not correct.
