Let $M := (e^{-aT_n}f(X_{T_n}); \mathcal{F}_n)$, where $a$ is constant, $\mathcal{F}_n$ is appropriately defined filtration and $T_n$ is $n$th jump of an independent (independent of $X_t$) Poisson process with parameter $\lambda$ i.e. $T_n$ is exponentially distributed with parameter $1/\lambda$.
Assume that $f(x) \geq \mathbb{E}_x[e^{-aU}f(X_{U})]$ for all $x \in \mathbb{R}_+$, where $U$ is independent of $X_t$ and exponentially distributed random variable with parameter $1/\lambda$.
Why does it follow from this inequality that $M$ is indeed a supermartingale? I assume this should be straightforward but I don't see how to apply the definition of a supermartingale here.
I think you're missing an assumption about $X$ being a stationary (strong) Markov process so that: $$\mathbb{E}[g(X_{t+u})|X_s, s\leq t] = \mathbb{E}_{X_t}[g(X_u)]$$
Also, if the $T_n$ are really jump times of a Poisson process, then the differences $T_{n+1}-T_n$ are exponentially distributed with parameter $1/\lambda$, obviously. I assume this was just a typo and not throwing you off.
With that said, because $T_n$ is $\mathcal{F}_n$-measurable, we can write: $$\mathbb{E}[M_{n+1}|\mathcal{F}_n] = e^{-aT_n}\mathbb{E}[e^{-a(T_{n+1}-T_n)}f(X_{T_{n+1}})|\mathcal{F}_n] = e^{-aT_n}\mathbb{E}[e^{-aU}f(X_{T_n+U})|\mathcal{F}_n]$$ for $U=T_{n+1}-T_n$. By the independence of $U$ and the stationary (strong) Markov property for $X$, we have: $$=e^{-aT_n}\mathbb{E}_{X_{T_n}}[e^{-aU}f(X_U)]$$ and by the hypothesized inequality:
$$\leq e^{-aT_n} f(X_{T_n}) = M_n$$