If $\left(X_n\right)_{n\in\mathbb{N}_0}$ is an $E$-valued stochastic process with distributions $\left(P_x\space:\space x\in E\right)$ satisfying $\mathrm{P}_x\left(X_0=x\right)=1$ and stochastic kernel $\kappa\left(x,B\right)=\mathrm{P}_x\left(X\in B\right)$, then the Markov property (namely "for every $A\in\mathcal{B}\left(E\right)$, every $x\in E$ and all $i,j\in\mathbb{N}_0$ we have $\mathrm{P}_x\left(\left.X_{i+j}\in A\space\right|\mathcal{F}_i\right)=\kappa_j\left(X_i,A\right)$") is implied if for every $A\in\mathcal{B}\left(E\right)$, every $x\in E$ and every $i\in\mathbb{N}_0$ we have $\mathrm{P}_x\left(\left.X_{i+1}\in A\space\right|\mathcal{F}_i\right)=\kappa_1\left(X_i, A\right)$ where $\kappa_j\left(x, A\right):=\mathrm{P}_x\left(X_j\in A\right)$ The following fact, easily derived from the given data, can be used without proof: $\kappa_{n+1}=\kappa_n\cdot\kappa_1:=\intop_E\kappa_n\left(\cdot, \mathrm{d}x\right)\kappa_1\left(x,\cdot\right)$ with $\kappa_0\left(x,\cdot\right):=\delta_x\left(\cdot\right)$, the Dirac measure.
This is Theorem 17.11 from Klenke's book (p. 350, Chapter 17: "Markov Chains")
Hint
If I understand correctly, all that needs be shown is that for all $m\in\mathbb{N}_0$, $\mathrm{P}_x\left(\left.X_{i+m}\in A\space\right|\mathcal{F}_i\right)=\kappa_m\left(X_i, A\right)$