Let $f\in C^2(\mathbb R)$ be positive and $h\ge 0$. Assume that $g:=f'/f$ is Lipschitz continuous and let $U$ be a strong solution of $${\rm d}U_t=\frac h2g(U_t){\rm d}t+\sqrt h{\rm d}W_t$$ ($W$ being a standard Brownian motion) with $U_0\sim\mu:=f\lambda^1$ (measure with density $f$ with respect to the Lebesgue measure $\lambda^1$). Let $$\pi_{s,\:t}(x,B):=\operatorname P\left[U_t\in B\mid U_s=x\right]\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R)$$ for $0\le s\le t$.
How can we show that $\mu$ is invariant with respect to $\pi_{s,\:t}$, i.e. $$\mu\pi_{s,\:t}=\mu\tag1$$ (where the left-hand side denotes the composition of transition kernels for all $0\le s\le t$?
It's clear to me that $$\operatorname P\left[U_s\in\:\cdot\:\right]\pi_{s,\:t}=\operatorname P\left[U_t\in\:\cdot\:\right]\;\;\;\text{for all }0\le s\le t\tag2.$$
Rescaling the process in time appropriately (which does not influence the invariant measures) and introducing $V$ such that $e^{-V} = f$, we can write: $${\rm d}U_t= - V'(U_t) {\rm d}t+\sqrt {2} \, {\rm d} B_t,$$ where $B_t$ is another Brownian motion. The associated Fokker-Planck equation, which governs the evolution of the probability density $p(u, t)$ of $U_t$, is $$ \frac{\partial p}{\partial t}(u, t) = \frac{\partial}{\partial u} \left(V'(u) \, p(u, t) + \frac{\partial p}{\partial u}(u, t) \right), $$ which admits $C \, e^{-V(u)}$ as a stationary solution. Note that the rescaling is not necessary, of course.