Can we show that a (random) time discretized Markov process is still Markov?

Question

Let

• $(E,\mathcal E)$ be a measurable space;
• $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$;
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
• $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$ and $$\mathcal F^{(t)}_n:=\mathcal F_{nt}\;\;\;\text{for }n\ge0$$ for $t\ge0$;
• $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname E\left[f(X_{s+t})\mid\mathcal F_s\right]=(\kappa_tf)(X_s)\tag1$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $s,t\ge0$ and $$X^{(t)}_n:=X_{nt}\;\;\;\text{for }n\ge0$$ for $t\ge0$;
• $\tau$ be a finite $(\mathcal F_t)_{t\ge0}$-stopping time on $(\Omega,\mathcal A)$.

Question: Are we able to show that $$\operatorname E\left[f\left(X^{(\tau)}_{m+n}\right)\mid\mathcal F^{(\tau)}_m\right]=(\kappa_\tau^nf)\left(X^{(\tau)}_m\right)\tag2$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $m,n\ge0$?

We can easily see that the claim is true when $\tau(\Omega)$ is countable. For the general case, let $\tau_k$ be a finite $(\mathcal F_t)_{t\ge0}$-stopping time on $(\Omega,\mathcal A)$ with $|\tau_k(\Omega)|\le\mathbb N$ for $k\in\mathbb N$ with $$\tau_k\ge\tau_{k+1}\;\;\;\text{for all }k\in\mathbb N\tag3$$ and $$\tau_k\xrightarrow{k\to\infty}\tau\tag4.$$ We clearly need to assume that $(X_t)_{t\ge0}$ is right-continuous in order to proceed. We now see that $$\mathcal F^{(\tau_k)}_m\supseteq\mathcal F^{(\tau)}_m\;\;\;\text{for all }m\ge0\text{ and }k\in\mathbb N\tag5$$ and $$X^{(\tau_k)}_n\xrightarrow{k\to\infty}X^{(\tau)}_n\;\;\;\text{for all }t\ge0.\tag6$$ By $(6)$ and the dominated convergence theorem, we see that $$\operatorname E\left[f\left(X^{(\tau_k)}_{m+n}\right)\mid\mathcal F^{(\tau)}_m\right]\xrightarrow{k\to\infty}\operatorname E\left[f\left(X^{(\tau)}_{m+n}\right)\mid\mathcal F^{(\tau)}_m\right]\tag7$$ for all $m,n\ge0$. On the other hand, by $(5)$ and $(2)$, we see that $$\operatorname E\left[f\left(X^{(\tau_k)}_{m+n}\right)\mid\mathcal F^{(\tau)}_m\right]=\operatorname E\left[\left(\kappa_{\tau_k}^nf\right)\left(X^{(\tau_k)}_m\right)\mid\mathcal F^{(\tau)}_m\right]\tag8$$ for all $m,n\ge0$ and $k\in\mathbb N$.

However, I don't know how to proceed with the right-hand side of $(8)$. I guess we need to assume some kind of continuity of $(\kappa_t)_{t\ge0}$ as well.