At first, I was under the impression that [itex]s_m[/itex] was constant. But shouldn't [itex]s_m[/itex] be inversely proportional to the distance?

The book never says this anywhere directly, but it seems to be implied by two other equations (for the intensity) given in the chapter:
$$I=\frac{P_{source}}{4\pi R^2}=\frac{1}{2}\rho v\omega ^2s_m^2$$
Wouldn't this mean that either the frequency or the amplitude decreases with distance? But the frequency can't change without a corresponding change in wavelength, so doesn't this mean the displacement-amplitude (and therefore also the pressure-amplitude) decreases with the distance?

I just want to make sure, because all the writing in the chapter seemed to imply to me that the amplitude was constant, yet these two equations for the intensity seem to say otherwise.

The wave provided is a plane wave solution, which does not decrease with distance travelled unless damped. It only describes a wave generated at a point source far away from the source (much further than the wavelength and the distance over which you are considering the wave).

Edit: Of course, if you solve the wave equation for a spherical wave, you will end up with a decreasing amplitude.

One more question; When you say that equation describes a planar wave, is this just because the amplitude is constant?

In other words, would the equation that describes a spherical wave be the same except with a variable amplitude?
(Something like [itex]\frac{c}{R}cos(kR-\omega t)[/itex]?)

Just as an exercise, you can do the following: Assume ##s(r,t) = f(r) \exp(i(kr - \omega t))## and insert into the wave equation (for r>0). The radial part of the Laplace operator is ##\partial_r^2 + (2/r) \partial_r##. Insert this into the wave equation and check that it solves it (again, for r > 0, you would need a source in r = 0 to keep it going).