Let $\sigma^2,\alpha,s,t>0$, $B\in\mathcal{B}(\mathbb{R})$ and $x\in\mathbb{R}$. Consider following integral
\begin{align}\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-\alpha t})}\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-\alpha s})}}\cdot\\ \int_\mathbb{R}\bigg[\int_B\exp\Big(&-\frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}\Big)dz\bigg]\quad\exp\Big(-\frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}\Big)dy\end{align}
I want to find it being equal to
$$\frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha(t+s)})}}\int_B \exp\Big(-\frac{(z-xe^{-\alpha (t+s)})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha(t+s)})}\Big)dz$$
which is a normal distribution. I think one can integrate over $B-ye^{-\alpha t}$ and then integrate by substitution, but I struggle finding the result. Thank you for any help!
Here is an approach that avoids directly manipulating the integrals (which is quite painful).
Let $$Y \sim N(xe^{-\alpha s}, \frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s}).$$ Let the conditional distribution of $Z$ given $Y=y$ be $$N(ye^{-\alpha t}, \frac{\sigma^2}{2\alpha} (1-e^{-2 \alpha t})).$$ Note that the joint distribution satisfies $f_{Y,Z}(y,z) = f_{Z \mid Y=y}(z) \cdot f_Y(y)$. This is precisely your integrand. In fact, the double integral is computing $P(Z \in B)$.
To compute this, one could instead find the marginal distribution of $Z$. I believe one can show that $(Y,Z)$ is bivariate Gaussian, so $Z$ is also Gaussian. Its expectation is $$E[Z] = E[E[Z \mid Y]] = E[Ye^{-\alpha t}] = xe^{-\alpha(s+t)}$$ and its variance is $$\text{Var}(Z) = E[\text{Var}(Z \mid Y)] + \text{Var}(E[Z \mid Y]) = \frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t}) + e^{-2\alpha t} \frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s}) = \frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)}).$$ Thus you can write down the density $f_Z$ of $Z$, and have $P(Z \in B) = \int_B f_Z(z) \, dz$. This is the goal integral.