Let
- $(E,\mathcal E)$ be a measurable space with $\{x\}\in\mathcal E$;
- $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$;
- $(\Omega,\mathcal A)$ be a measurable space;
- $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$;
- $(\mathcal F^X_t)_{t\ge0}$ denote the filtration generated by $(X_t)_{t\ge0}$;
- $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ with $$\operatorname P_x[X_0=x]=1\tag1$$ and $$\operatorname E_x\left[f(X_{s+t})\mid\mathcal F^X_s\right]=(\kappa_t)(X_s)\tag2$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $s,t\ge0$ for $x\in E$;
- $\tau$ be an $(\mathcal F^X_t)_{t\ge0}$-stopping time;
- $\mu$ be a probability measure on $(E,\mathcal E)$ with $$\int\mu({\rm d}x)\kappa_t(x,B)=\mu(B)\tag3\;\;\;\text{for all }B\in\mathcal E)$$ for all $t\ge0$.
Does the invariance property $(3)$ allow to simplify $$\int\mu({\rm d}x)\operatorname E_x\left[f(X_t)1_{\{\:t\:<\:\tau\:\}}\right]\tag4$$ for bounded $\mathcal E$-measurable $f:E\to\mathbb R$.
Clearly, $$\int\mu({\rm d}x)\operatorname E_x\left[f(X_t)\right]=\int f\:{\rm d}\mu.\tag5$$ So, the "problem" is $\{t<\tau\}\in\mathcal F^X_t$.
No, not for general stopping times, because you need to take path measures/expectations to check whether the stopping time is after $t$ or not.
Maybe you'd get some use from applying the weak Markov property:
\begin{align*} \mathbb{E}_x\left[f(X_t)1_{\{t < \tau\}}\right] &= \mathbb{E}_x\left[1_{\{t < \tau\}}\mathbb{E}_x \left[ f(X_1) \circ \theta_{t-1} \mid \mathcal{F}_{t-1} \right]\right] \\ &= \mathbb{E}_x\left[1_{\{t < \tau\}}\mathbb{E}_{X_{t-1}} \left[ f(X_1) \right]\right] \\ &= \mathbb{E}_x\left[1_{\{t < \tau\}}\kappa f (X_{t-1})\right]. \end{align*}