Let $f$ be a bounded and measurable function. Then
$|U^\alpha f(x)| \leq \mathbb{E}^x \left(\int_0^\infty e^{-\alpha t} |f(X_t)| \,dt \right) \leq \mathbb{E}^x \left(\int_0^\infty e^{-\alpha \cdot t} \cdot \|f\|_\infty \, dt \right) = \|f\|_\infty \cdot \int_0^\infty e^{-\alpha \cdot t} \, dt < \infty$
and therefore the function $x \mapsto U^\alpha f(x)$ is still a bounded (and measurable) function. Thus
$ (U^\beta f)(x)-(U^\alpha f)(x) = \mathbb{E}^x \left(\int_0^{\infty} (e^{-\beta \cdot t}-e^{-\alpha \cdot t}) \cdot f(X_t) \, dt \right) $
and
$(U^\beta (U^\alpha g))(x) = \mathbb{E}^x \left( \int_0^{\infty} e^{-\beta \cdot t} \cdot(U^\alpha g)(X_t) \, dt \right)$
Iterating the last equation you obtain an expression for $(U^\alpha)^n$.
To show that $U^\alpha$ fulfills the resolvent equation you have to know that $U^\alpha = (\alpha \cdot \text{id}- A)^{-1}$ where $A$ is the generator (so you especially have to know some more stuff about generators of semigroups). If you know this, the equation follows like this:
$U^\beta f-U^\alpha f = U^\alpha((\alpha \cdot \text{id}-A) U^\beta-\underbrace{\text{id}}_{(\beta \cdot \text{id}-A)U^\beta}) f = U^\alpha(\alpha-\beta)U^\beta f = (\alpha-\beta)U^\alpha U^\beta f$