Think of the operator

$$Tf(x)=2\int_0^\infty \frac{e^{-\frac{x^2}{2z}}}{\sqrt{2\pi z}} f(z)dz$$

as an integral operator on the space $L^2(e^x dx)$ of functions with $\int_{-\infty}^\infty |f(x)|^2 dx <\infty$. That is

$$Tf(x)=\int_{-\infty}^\infty K(x,y) e^y dy$$

with $K(x,y)= 2I[y>0] \frac{e^{-\frac{x^2}{2y} -y}}{\sqrt{2\pi y}}$.  It is easy to verify that

$$\int\int K(x,y)^2 e^{x+y}dxdy <\infty$$

so $K$ is Hilbert-Schmidt on $L^2(e^x dx)$, hence compact. Because $K(x,y)>0$ for all $x,y$ the Perron-Frobenius theorem (suitably generalized to compact operators of this type) shows that $T$ has a unique positive eigenvalue $\lambda_0$ with a positive eigenfunction and all other eigenvalues $\lambda$ are of modulus $|\lambda|<\lambda_0$. I claim that $T\phi=\phi$ where $\phi(x)=e^{-2|x|}$, so the unique positive eigenvalue is one! (Certainly this has to do with the origins of the problem in probability theory.)   This explains the identity up to an overall constant on the r.h.s..  The constant is given by the inner product of 

$$\frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}$$

and the left eigenvector of $T$ with eigenvalue $1$.  This must be one, but I don't see why.

To see that $T\phi=\phi$ it is most convenient to take a Fourier transform in $x$ to get

$$ \widehat{T\phi}(k)=2 \int_0^\infty e^{-\frac{k^2}{2}z}e^{-2z}dz=\frac{4}{k^2 +4}=\widehat{\phi}(k).$$