Verify that $-iH$ is the generator of $U_t$, $t\in\R$. Make an educated guess for an integral formula for $U_t$ in case of the Poisson and the Ornstein-Uhlenbeck semigroup.
\begin{eqnarray*}
T(z)f(x)&=&\int_{S^{d-1}}\frac{(1-z^2)}{\norm{zx-y}^d}\,f(y)\,\s(dy)\\
T(z)f(x)&=&\Big(\frac{1-z^2}{2\pi}\Big)^{d/2}
\int\exp\Big(-\frac{\norm{zx-y}^2}{2(1-z^2)}\Big)f(y)\,dy
\end{eqnarray*}
and put $z=e^{-it}$. A priori it's not obvious how to interpret the norm $\norm{zx-y}^2$: it might be
$$
\sum_{j=1}^d(zx_j-y_j)^2
\quad\mbox{or}\quad
\sum_{j=1}^d|zx_j-y_j|^2
$$
If you want $z\mapsto T(z)f$ to be analytic you'll probably choose the first interpretation. However in this case the kernel turns to be oscillatory.