Let $(X_t)$ be a Markov process with respect to a filtration $\mathcal{F}_t$. Assume that $P(X_t>0 \, \forall t\geq 0) = 1 $.
Denote $E_x$ the expectation under the measure where $X_0=x$.
Is it true that $$ E_x[X_{t+s}^{-1} e^{z X_{t+s}} \rvert \mathcal{F_t}] = E_{X_t}[X_{s}^{-1} e^{z X_{s}} ]? $$ for $z\in \{z \in \mathbb{C} : \Re(z) <0 \} $ given both expressions exist.
My guess is yes, since $f_n(x) = x^{-1} e^{z x } 1_{\{x>1/n\}}$ is bounded and I can do $$ E_x[ f_n(X_{t+s} ) \rvert \mathcal{F_t}] = E_{X_t}[ f_n(X_s) ] $$ for all $n$ and take the limit. Does that work?