Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(E,\mathcal E)$ be a measurable space, $(W_n)_{n\in\mathbb N_0}$ be a time-homogeneosu Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ be a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\lambda>0$ independent of $W$.

I would like to conclude that $$X_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with transition semigroup$^1$ $\left(e^{t(\kappa-\lambda)}\right)_{t\ge0}$.

The idea is \begin{equation}\begin{split}\operatorname P\left[X_{s+t}\in B\mid\mathcal F^X_s\right]&=\operatorname P\left[Y_{N_s+(N_{s+t}-N_s)}\in B\mid\mathcal F^X_s\right]\\&=\sum_{n=0}^\infty\operatorname P\left[N_{s+t}-N_s=n,Y_{N_s+n}\in B\mid\mathcal F^X_s\right]\\&=\sum_{n=0}^\infty\operatorname P\left[N_{s+t}-N_s=n\right]\operatorname P\left[Y_{N_s+n}\in B\mid\mathcal F^X_s\right]\\&=e^{-\lambda t}\sum_{n=0}^\infty\frac{(\lambda t)^n}{n!}\operatorname P\left[Y_{N_s+n}\in B\mid\mathcal F^X_s\right].\end{split}\tag1\end{equation} for all $B\in\mathcal E$ and $s,t\ge0$.

However, in order for the third equality in $(1)$ to hold, we need that $N_{s+t}-N_s$ is independent of $\mathcal F^X_s$ for all $s,t\ge0$.

Intuitively, this seems to be obvious, since $(N_t)_{t\ge0}$ is independent of $(W_n)_{n\in\mathbb N}$ and $N_{s+t}-N_s$ is independent of $\mathcal F^N_s$ for all $s,t\ge0$, but how can we prove it rigorously?

It's easy to see that $$\left.\sigma(X_s)\right|_{\{\:N_s\:=\:n\:\}}=\left.\sigma(W_n)\right|_{\{\:N_s\:=\:n\:\}}\tag3$$ for all $s\ge0$ and $n\in\mathbb N$. So, maybe the desired claim follows from the local property of conditional expectation.

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(E,\mathcal E)$ be a measurable space, $(W_n)_{n\in\mathbb N_0}$ be a time-homogeneosu Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ be a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\lambda>0$ independent of $W$.

I would like to conclude that $$X_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with transition semigroup$^1$ $\left(e^{t(\kappa-\lambda)}\right)_{t\ge0}$.

The idea is \begin{equation}\begin{split}\operatorname P\left[X_{s+t}\in B\mid\mathcal F^X_s\right]&=\operatorname P\left[Y_{N_s+(N_{s+t}-N_s)}\in B\mid\mathcal F^X_s\right]\\&=\sum_{n=0}^\infty\operatorname P\left[N_{s+t}-N_s=n,Y_{N_s+n}\in B\mid\mathcal F^X_s\right]\\&=\sum_{n=0}^\infty\operatorname P\left[N_{s+t}-N_s=n\right]\operatorname P\left[Y_{N_s+n}\in B\mid\mathcal F^X_s\right]\\&=e^{-\lambda t}\sum_{n=0}^\infty\frac{(\lambda t)^n}{n!}\operatorname P\left[Y_{N_s+n}\in B\mid\mathcal F^X_s\right].\end{split}\tag1\end{equation} for all $B\in\mathcal E$ and $s,t\ge0$.

However, in order for the third equality in $(1)$ to hold, we need that $N_{s+t}-N_s$ is independent of $\mathcal F^X_s$ for all $s,t\ge0$.

Intuitively, this seems to be obvious, since $(N_t)_{t\ge0}$ is independent of $(W_n)_{n\in\mathbb N}$ and $N_{s+t}-N_s$ is independent of $\mathcal F^N_s$ for all $s,t\ge0$, but how can we prove it rigorously?

It's easy to see that $$\left.\sigma(X_s)\right|_{\{\:N_s\:=\:n\:\}}=\left.\sigma(W_n)\right|_{\{\:N_s\:=\:n\:\}}\tag3$$ for all $s\ge0$ and $n\in\mathbb N$. So, maybe the desired claim follows from the local property of conditional expectation.

EDIT: Let $$\mathcal E_t:=\bigcup_{n\in\mathbb N_0}\left\{A_1\cap\left\{N_t=n\right\}\cap A_2:n\in\mathbb N\text{ and }(A_1,A_2)\in\mathcal F^N_t\times\mathcal F^W_n\right\}$$ for $t\ge0$. It's easy to see that

1. $\emptyset\in\mathcal E_t$ for all $t\ge0$;
2. $\mathcal E_t$ is closed under finite intersections for all $t\ge0$;
3. $N_{s+t}-N_s$ is independent of $\mathcal E_s$ for all $s,t\ge0$.

All these properties together imply that $N_{s+t}-N_s$ is independent of $\sigma(\mathcal E_s)$ for all $s,t\ge0$. So, it would be sufficient to show that $\sigma(\mathcal E_s)=\mathcal F^X_s$ for all $s\ge0$.