Do modern Physicists really think about Electromagnetic Waves in this
way? Is this kind of propagation "mechanism" really even needed?
I expect there is variability among all the people who study Maxwell's equations. Here's my take. The equations and all the quantities they contain are mathematical abstractions. We use these abstractions in prescribed ways to come up with some predictions for field values, powers, etc. We map these predictions to observable, measurable phenomena like resistors heating up or needles flicking on a dial. How you get these predictions really doesn't depend on how you think about the details of propagation. The solutions to Maxwell's equations are, regardless of how you think about them. We then go out and do some experiments to see if our predictions matched our measurements. When we do this, we find that Maxwell's equations are good predictors of every observed classical EM phenomenon. In this context, questions about the nature of $E$ and $B$ "creating" each other become kind of meaningless. What we have is a theory that describes experiment when we use it. How we choose to interpret the details of the theory don't change it's predictions; the only benefits to thinking about it one particular way or another is to help us intuit, recall, explain, or reason about the theory. In those areas, I find the idea that "$E$ begets $B$ begets $E$ begets $B$ begets ..." can be useful for recalling the two curl equations and how they combine into wave equations, or for how the Yee grid FDTD method works. But if you ask me if this is what is "really" happening when I flip on the light switch, my answer is "I can predict many useful things about the light when the switch is flipped without knowing the answer to your question."
I will be going to set $\epsilon_0=\mu_0=1$. Now the Maxwell equations are:
$$\nabla . \textbf{E} = \rho$$
$$\nabla . \textbf{B} = 0$$
$$\nabla \times \textbf{E} = -\frac{\partial\textbf{B}}{\partial t}$$
$$\nabla \times \textbf{B} = \left(\textbf{J} + \frac{\partial\textbf{E}}{\partial t}\right)$$
and we have the identity
\begin{equation}
\nabla \times (\nabla \times\textbf{V}) = \nabla(\nabla.\textbf{V}) - \nabla^2\textbf{V}
\end{equation}
Now proceeding in the same fashion as in the vacuum
$$\nabla \times (\nabla \times \textbf{E}) = \nabla \times -\frac{\partial\textbf{B}}{\partial t} = -\frac{\partial}{\partial t}(\nabla \times \textbf{B})=-\frac{\partial}{\partial t}\left(\textbf{J} + \frac{\partial\textbf{E}}{\partial t}\right)$$
while the LHS becomes:
$$\nabla(\nabla.\textbf{E}) - \nabla^2\textbf{E}=\nabla(\rho) - \nabla^2\textbf{E}$$
Rearranging RHS and LHS we get
$$\nabla^2\textbf{E}-\frac{\partial^2\textbf{E}}{\partial t^2}=\nabla\rho +\frac{\partial}{\partial t}\textbf{J}$$
In simpler terms $$\Box\textbf{E}=\textbf{C}$$
where $$\textbf{C}=\nabla\rho +\frac{\partial}{\partial t}\textbf{J}$$
Now moving to the case of $\textbf{B}$
$$\nabla \times (\nabla \times \textbf{B})=\nabla \times\left(\textbf{J} + \frac{\partial\textbf{E}}{\partial t}\right)= \nabla \times\textbf{J} + \frac{\partial}{\partial t}(\nabla\times\textbf{E})=\nabla \times\textbf{J} -\frac{\partial^2\textbf{E}}{\partial t^2}$$
as for LHS we have
$$\nabla(\nabla.\textbf{B}) - \nabla^2\textbf{B}=\nabla(0) - \nabla^2\textbf{B}$$
Rearranging RHS and LHS we get
$$\nabla^2\textbf{B}-\frac{\partial^2\textbf{B}}{\partial t^2}=-\nabla \times\textbf{J} $$
in simpler terms $$\Box \textbf{B}=\textbf{F}$$
where $$\textbf{F}=-\nabla \times\textbf{J}$$
So putting sources has ultimately lead to what we call inhomogeneous wave equation which is simply $$\Box f(t,\vec{x})=h(t,\vec{x})$$ same thing as in case of Laplacian and Poisson equation in chapter 3.
Bonus material(I will make an assumption of tensors):
Maxwell equations are Lorentz covariant equation (that's how they contributed in Einstein triumph of special relativity), even when they were discovered in the era of Newtonian mechanics. Lorentz covariance another term to say a given physical quantity obey transformation law of different inertial reference frames in special relativity.
You might have also noticed how messy it becomes taking curl and div each time in above calculation and you will see it when you compare equation of chapters 10 and 12 of Griffiths book relating $\vec{J},\rho, A_\mu$. I will provide a rough sketch of the above calculation in the light of SR.
We define a quantity called 4-vector a generalization of vectors in 4 dimensions of Minkowski space $$A_{\mu}=(V, A_x, A_y, A_z)$$
$$J_{\mu}=(\rho, J_x, J_y, J_z)$$
Define a quantity called electromagnetic strength tensor $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$
Maxwell equation can be recast as $$\partial^\nu F_{\mu\nu}=J_\mu$$ and $$\partial_{[\mu} F_{\nu\lambda]}=0$$
Leave the second equation aside (it's actually a tautology) let's focus on first equation expanding it in terms of $A_\mu$
$$\partial^\nu(\partial_\mu A_\nu-\partial_\nu A_\mu)=J_\mu$$
$$\partial^\nu(\partial_\mu A_\nu)-\partial^\nu(\partial_\nu A_\mu)=J_\mu$$ rearanging the terms we have
$$\partial_\mu(\partial^\nu A_\nu)-(\partial^\nu\partial_\nu) A_\mu=J_\mu$$
Now we use the Lorentz gauge and set $\partial^\nu A_\nu$ so ultimately we're left with
$$-(\partial^\nu\partial_\nu) A_\mu=J_\mu$$ which is nothing but $$-\Box A_\mu = J_\mu$$
which is simply the wave equation of different potentials under the presence of different sources you can recover $\vec{E}$, $\vec{B}$ from $A_\mu$. You might not have followed anything from this bonus material if it's your first encounter of,4-vectors, tensors, Einstein summation, Gauge transformation/freedom. What I actually wanted to show you was bear for the time being in the complicated mess of calculation and when you're done with chapter 12 of Griffith you'll have a different outlook at Electrodynamics as a whole.
Best Answer
I think the answer is simply: "Yes".
What you should keep in mind is energy conservation: As long as there are no sources, the total energy of the electromagnetic field is conserved.
A source, which is possibly localized somewhere and not necessarily non-zero at all times.