Let
- $(E,\mathcal E)$ be a measurable space;
- $\mathcal E_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$;
- $(\Omega,\mathcal A)$ be a measurable space;
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
- $(X^1_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued $(\mathcal F_t)_{t\ge0}$-adapted process on $(\Omega,\mathcal A)$;
- $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ with $\operatorname P_x[X^1_0=x]=1$ and $$(Q_tf)(x):=\operatorname E_x\left[f(X^1_t)\right]\;\;\;\text{for }f\in\mathcal E_b\text{ and }t\ge0$$ for $x\in E$;
- $\rho:E\to[0,\infty)$ be $\mathcal E$-measurable and $$M_t:=\exp\left(-\int_0^t\rho(X^1_s)\:{\rm d}s\right)\;\;\;\text{for }t\ge0.$$
- $\xi$ be a real-valued random variable on $(\Omega,\mathcal A)$ such that $\xi$ is $\operatorname P_x$-independent of $X^1$ and $\operatorname P_x\circ\xi^{-1}$ is the exponential distribution with parameter $1$ for all $x\in E$;
- $$\tau_1:=\inf\left\{t\ge0:\int_0^t\rho(X^1_s)\:{\rm d}s\ge\xi\right\}.$$
We can show that $$\operatorname P_x\left[\tau_1\le t\right]=\operatorname E\left[\int_0^tM_s\rho(X_s)\:{\rm d}s\right]\;\;\;\text{for all }t\ge0\tag1.$$ for all $x\in E$. Now let $(X^2_t)_{t\ge0}$ be an independent copy of $X^1$ and $$X_t:=\left.\begin{cases}X^1_t&\text{, if }t<\tau_1\\X^2_{t-\tau_1}&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }t\ge0.$$
What I've done: Let $t\ge0$ and $(x,B)\in E\times\mathcal E^{\otimes[0,\:\infty)}$. Clearly, $$\operatorname P\left[X_t\in B;t<\tau_1\right]=\operatorname P\left[X^1_t\in B;t<\tau_1\right]\tag2.$$ Now let $$C:=\left\{(s,x)\in[0,t]\times E^{[0,\:\infty)}:x(t-s)\in B\right\}$$ and $$C_x:=\left\{s\in[0,t]:(s,x)\in C\right\}\;\;\;\text{for }x\in E^{[0,\:\infty)}.$$ Since $X^1$ and $X^2$ are $\operatorname P_x$-independent, $\tau_1$ and $X^2$ are $\operatorname P_x$-independent as well. Thus, using $(1)$, \begin{equation}\begin{split}\operatorname P\left[Y_t\in B;t\ge\tau_1\right]&=\int\operatorname P\left[X^2\in{\rm d}x\right]\operatorname P\left[\tau_1\in C_x\right]\\&=\int\operatorname P\left[X^2\in{\rm d}x\right]\operatorname E\left[\int_{C_x}M_s\rho(X_s)\:{\rm d}s\right].\end{split}\tag3\end{equation} But is this helpful?
Another thing we might note is $$\operatorname E_\mu\left[f\left(X^2_{t-\tau_1}\right);\tau_1\le t\mid\mathcal F_{s-\tau_1}\right]=1_{\left\{\:\tau_1\:\le\:s\:\right\}}\left(\kappa_{t-\tau_1}f\right)(X_s)\tag4$$ for all $t\ge s\ge0$, where $$\operatorname P_\mu[B]:=\int\mu({\rm d}x)\operatorname P_x\left[B\right]\;\;\;\text{for }(x,B)\in E\times\mathcal E,$$ for every probability measure $\mu$ on $(E,\mathcal E)$.
How can we determine the transition semigroup of $X$?
Remark: Please take note of my related question on mathoverflow.