Given an SDE in a Hilbert Space $H$ and an underlying probability space $(\Omega, \mathcal{F}, P)$ with solution $(X_t)_{t\ge 0}$ and writing $X_t^x$ as the solution with initial condition $x$, define the following semigroup $(P_t)$ on the space of bounded Borel-measurable functions on $H$ via $$ (P_t\varphi)(x) = \mathbb{E}[ \varphi(X_t^x)]. $$ Next, define a semigroup $(P_t^*)$ on the space of all probability measures on $H$ via $$ (P_t^*\mu)(\varphi) = \mu(P_t\varphi), $$ where we use the notation $\mu(f) = \int_H f(x)\,d\mu(x)$.
Now for my question, if $X_t$ has law $\nu_t$, then the claim is that $P^*_{t}\nu_s = \nu_{t+s}$. I'm having trouble unwinding all of the definitions to see why this is true. But moreover, this seems like a standard procedure in defining semigroups. I'm not too familiar with subject, but is this somehow related to transition probabilities of a Markov process?
There are a lot of little things going on here that I'm somewhat familiar with, but I'm having a bit of trouble synthesizing to see the big picture.
I ended up figuring it out. We show it for $s=0$ and then use the semigroup property to conclude. First, by changing variables, we note if $X$ has distribution $\mu$, then with my notation above, $\mu(f) = E[f(X)]$. With this, for $f\in C_b(H)$ (i.e. continuous bounded functions on $H$), we see $$ (P^*_tf)(\nu_0) = \nu_0(P_tf)=E[(P_tf)(X_0)] = (P_0(P_tf))=P_tf=E[f(X_t)] = \nu_t(f), $$ which gives $P_t^* \nu_0 = \nu_t$.
We then conclude $P^*_t \nu_s = P^*_tP^*_s \nu_0 = P^*_{t+s}\nu_0 = \nu_{t+s}$.