How can we conclude from Maxwell's wave equation that the speed of light is the same regardless of the state of motion of the observers?

Your question is an excellent one and you are right about the $\nabla$ operator. And you are also right about the insufficiency of the argument you report in the book you are reading.

To make the argument more carefully, there are two options. The first would be to work out how the Maxwell equations themselves change as you go to another inertial frame. That would take a lot of calculating if you start from first principles. (And by the way, they don't change---you get back the same equations but now in terms of ${\bf E}', {\bf B}', \rho', {\bf j}', {\bf \nabla}', \partial/\partial t'$).

A second option, mathematically easier but still requiring some work if you are not familiar with it, is to show that the $\nabla$ operator and the $\partial/\partial t$ operator have a special property: when you combine them in the combination $$ \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} $$ then their effect is the same as $$ \nabla'^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t'^2} $$ All the changes when moving from unprimed to primed coordinates cancel out. If you are familiar with partial differentiation then you could try checking this. When you learn the subject more fully, it becomes an example that can be handled more easily using the language of 4-vectors.

I think that McMahon might possibly have not thought carefully enough about what he was deriving and what he was assuming in his argument. He might for example have been taking it for granted that the Maxwell equations themselves take the same form in all inertial frames. But if he did not first prove that in his book then he ought not to claim that the derivation of waves of given speed from them proves that the wave speed will be independent of the motion of the source.

If Maxwell's equations have the same form in all frames of reference, then the wave speed is defined by the product of two physical constants, irrespective of coordinate system. i.e. Your book just implicitly assumes that, but of course it requires experimental testing - i.e. Michelson-Morley etc.

Your observation is correct, Maxwell's equations alone do not imply an invariant speed of light. One can make a Galilean transformation and get an observer-dependent speed of light as shown in the answer to this question. However, the derivation of Maxwell's equations makes no assumption of a privileged reference frame: $\varepsilon_0$ and $\mu_0$ are assumed to be properties of the vacuum. Yes, a coordinate system must be chosen, but from the point of view of derivation of the equations this is totally arbitrary. In order to keep a non-constant speed of light one would have to retroactively make the assumption after the fact that the coordinates chosen happened to be stationary coordinates with respect to the aether.