The wave operator decomposes as
$$
\partial_t^2-\Delta = (\partial_t-i\sqrt{-\Delta})(\partial_t+i\sqrt{-\Delta}),
$$
so you can think of $e^{it\sqrt{-\Delta}}$ as solving a "half of" the wave equation. The dispersion relation is $\omega = |\xi|$, where $\omega$ and $\xi$ are the Fourier variables for $t$ and $x$, respectively. On the other hand, the Schrödinger propagator $e^{it\Delta}$ has the dispersions relation $\omega=|\xi|^2$, which makes it genuinely dispersive, i.e., the propagation speed depends on the frequency.

Note that $e^{it\Delta}$ is *not* the heat semigroup, which the other answers and comments seem to suggest.