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

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

The expression you are interested in is of the form $\lim_n T^n\psi_t$ where $T$ is the integral operator $$Tf(x)=2\int_0^\infty \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2\pi y}} f(y)dy$$ and $\psi_t(x)= \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}$. Note that $T$ is an operator with a positive kernel $T(x,y)=2 I[y>0] \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2\pi y}}$ which satisfies $\int_{-\infty}^\infty T(x,y)dx =1.$ That is, $T$ is much like a stochastic matrix and the limit you wish to obtain could only hold if $e^{-2|x|}$ is the principle eigenvector (with eigenvalue $1$).

To realize $T$ as something like a stochastic matrix, we should present it as a compact operator. It is not compact on $L^2(\mathbb{R})$, but if we think of it as an integral operator on the space $L^2(e^x dx)$ of functions with $$\int_{-\infty}^\infty |f(x)|^2 dx <\infty,$$ i.e. $$Tf(x)=\int_{-\infty}^\infty K(x,y) f(y) e^y dy$$ with $K(x,y)= 2I[y>0] \frac{e^{-\frac{x^2}{2y} -y}}{\sqrt{2\pi y}}$, then $\int\int K(x,y)^2 e^{x+y}dxdy <\infty$ so $T$ 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.) To see that $T\phi=\phi$ it is most convenient to take a Fourier transform in $x$ to get $$ \widehat{T\phi}(k)=2 \int_0^\infty e^{-\frac{k^2}{2}z}e^{-2z}dz=\frac{4}{k^2 +4}=\widehat{\phi}(k).$$

The other thing we need is the left principle eigenvector -- the eigenvector of $T^\dagger$ with eigenvalue $1$. Here the adjoint must be on $L^2(e^x dx)$ so

$$ T^\dagger f(x) =\int_{-\infty}^\infty K(y,x)f(y)e^y dy= 2 e^{-x} I[x>0] \int_{-\infty}^\infty \frac{e^{-\frac{y^2}{2x}}}{\sqrt{2\pi x}} e^y f(y).$$

Observe that $T^\dagger \widetilde{\phi}(x)=\widetilde{\phi}(x)$ where $\widetilde{\phi}(x)=2 e^{-x}I[x>0]$. The factor of $2$ enforces the normalization $\langle \phi,\widetilde \phi \rangle =1$ (with the inner product in $L^2(e^xdx)$). It now follows that

$$T^n f =\phi \langle \widetilde{\phi},f\rangle +o(1)$$

for any function $f\in L^2(e^xdx)$. Since

$$\langle \widetilde{\phi},\psi_t \rangle =\int_{0}^\infty \widetilde{\phi}(x)\psi_t(x)e^x dx=2 \int_0^\infty \psi_t(x)= $$

your identity follows.