= Solution
The first condition is that $X$ is Markov with transition semigroup
$P(t)=e^{tQ}$. The second is that, for every function
$f:S\to\mathbb R$,
$$
M_t^f=f(X_t)-f(X_0)-\int_0^t(Qf)(X_s)\,ds
$$
is a martingale.
The first condition implies the second by conditioning over a short time interval, using
$P(h)f=f+hQf+o(h)$, summing increments, and passing to the limit. This is Dynkin's formula.
Conversely assume the <martingale problem for a continuous-time Markov chain>. Fix $T>0$ and a function $f$, and put
$$
u(t,x)=(e^{(T-t)Q}f)(x).
$$
The backward equation gives $\partial_tu+Qu=0$. Applying the martingale identity to this time-dependent function shows that $u(t,X_t)$ is a martingale. Therefore, for $s\leq T$,
$$
\mathbb E[f(X_T)\mid\mathcal F_s]
=u(s,X_s)=(e^{(T-s)Q}f)(X_s).
$$
Taking indicator functions $f$ proves both the Markov property and the transition semigroup $e^{tQ}$.
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