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.