Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
- $(E,\mathcal E)$ be a measurable space;
- $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued $(\mathcal F_t)_{t\ge0}$-adapted process on $(\Omega,\mathcal A,\operatorname P)$;
- $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$.
Suppose we want to show that $(X_t)_{t\ge0}$ is $(\mathcal F_t)_{t\ge0}$-Markov with transition semigroup $(\kappa_t)_{t\ge0}$; i.e. $$\operatorname E\left[f(X_{s+t})\mid\mathcal F_s\right]=(\kappa_tf)(X_s)\tag1$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $s,t\ge0$.
Why is it sufficient to show that, for fixed $s\ge0$, $F\in\mathcal F_s$ and bounded $\mathcal E$-measurable $f:E\to\mathbb R$, the measures $\mu$ and $\nu$ on $\mathcal B([0,\infty))$, defined by \begin{align}\mu((a,b])&:=\operatorname E\left[1_F\int_a^bf(X_{s+t})\:{\rm d}t\right];\\\nu((a,b])&:=\operatorname E\left[1_F\int_a^b(\kappa_tf)(X_s)\:{\rm d}t\right]\end{align} for $0\le a<b$, coincide?
We may note that the Laplace transform of $\nu$ is given by \begin{equation}\begin{split}\mathcal L_\nu(\lambda)&=\int\nu({\rm d}t)e^{-t\lambda}\\&=\operatorname E\left[1_F\int_0^\infty e^{-t\lambda}(\kappa_tf)(X_s)\right]\\&=\operatorname E\left[1_F\left(\tilde R_\lambda f\right)(X_s)\right],\end{split}\tag2\end{equation} where $$\tilde R_\lambda f:=\int_0^\infty e^{-t\lambda}\kappa_tf\:{\rm d}t$$ for all $\lambda\ge0$.
Most probably we need further assumptions, but ignoring rigor for a moment, we may note the following: If $(\mathcal D(A),A)$ denotes the infinitesimal generator of $(\kappa_t)_{t\ge0}$ (considered as a (contractive) semigroup on the space of bounded $\mathcal E$-measurable (maybe even continuous; I don't know if this matters) functions $E\to\mathbb R$ equipped with the supremum norm), we know (from general semigroup theory) that the resolvent $R_\lambda(A)$ for $\lambda\in\rho(A)$ is equal to $\tilde R_\lambda$ (see Theorem 2.12 here and note that we can take $\omega=0$).
Unless mistaken from the definition of $\mu$ and $\nu$ we have for any $0<a<b$ : $$\mu((a,b])=E\left[1_F\int_a^bf(X_{s+t})\:{\rm d}t\right]=E\left[1_F.E\left[\int_a^bf(X_{s+t})\:{\rm d}t|\mathcal{F}_s\right]\right]=\nu[(a,b])= E\left[1_F\int_a^b(\kappa_tf)(X_s)\:{\rm d}t\right]$$
Now $(\kappa_tf)(X_s)$ is $\mathcal{F}_s$ measurable and from a.s. unicity of conditional expectation with respect to $\mathcal{F}_s$ you get almost surely : $$\operatorname E\left[f(X_{s+t})\mid\mathcal F_s\right]=(\kappa_tf)(X_s)$$ Now you can finish the argument as the interval $(a,b]$ is indeed a $\pi$-system as suggested by William M. so that it's true for $\sigma((a,b] b>a>0)= \mathcal B(\mathbb R^+)$ (after completion unless mistaken)