Relativistic momentum of the E.M field vs Poynting momentum questions I come somehow to the following thoughts:
The energy of the EM field is $$\mathcal{E} = {\epsilon_0\over 2} (E^2 + c^2 B^2).$$
Associate to $\mathcal E$ the relativistic mass $$m_r = {\mathcal{E}\over c^2}$$ and the
relativistic momentum $${\mathbf P} = m_r {\mathbf v} = {\mathcal{E} \mathbf v\over c^2}.$$
On the other hand, the Poynting momentum of the E.M. field is
$${\mathbf P} = \epsilon_0(\mathbf E \times \mathbf B)$$
Equating, we find
$${1\over 2c^2} (E^2 + c^2 B^2)\mathbf v = \mathbf E \times \mathbf B.$$
Raising to the square:
$${1\over 4c^4} (E^2 + c^2 B^2)^2 \mathbf v^2 = ||\mathbf E||^2||\mathbf B||^2 - \mathbf E\cdot \mathbf B,$$
which gives $\mathbf v$ in function of $\mathbf E$ and $\mathbf B$.
Now, let find at what conditions $v = ||\mathbf v||$ is equal to $c$.
From the previous equation, with $v = c$, we have
$$E^4 + c^4B^4 + 2c^2 E^2B^2 = 4 c^2 E^2 B^2 -  4c^2 \mathbf E\cdot \mathbf B.$$
Hence
$$(E^2 - c^2B^2)^2 = -4c^2\mathbf E\cdot \mathbf B.$$
We see that for this equation to be possible, $\mathbf E$ must be orthogonal to $\mathbf B$ and
$E$ must be equal to $c B$.
It is not difficult, by superimposing two waves of this form but propagating at right angle, to
generate an E.M. field which does not fulfill these conditions. That would mean that inside
the interference zone, the speed of the energy propagation is lesser than $c$.
First question: for mechanical waves, is there an analog to this phenomenon? that is, by superimposing two
mechanical waves, can you create a situation where the energy propagates slower than the natural speed of the wave inside the given medium (assuming such an energy propagation has been defined somehow, which I believe is possible).
Second question: Trying to contradict the above derivation, I try to imagine a situation where an EM wave
is propagating into free space, inside a waveguide for example, and somehow, the E field is not orthogonal
to the B field. By propagating, I mean that the EM field is null at every time $< r/c$, where $r$ is the distance of observation from the source. This would destroy the above deduction, since obviously, the energy of such a wave must propagate at speed $c$. My question is: could such a wave be generated?
EDIT: actually, the above equation only implies $\mathbf E\cdot \mathbf B < 0$, that is, the E-field makes an obtuse angle with the B-field. This does not change to much the nature of the problem and the questions. So, the last question becomes: is it possible to create a propagating EM wave inside empty space such that the E-field and B-field make an acute angle ?
 A: A couple of things to think about.
You are looking at this in a vacuum without boundary conditions. So E and B will be orthogonal and the phase and group velocities will be the same, and the energy flow will be perpendicular to the wave fronts. Also you are not considering turning the wave source on and off.
As soon as you have an aperture, waveguide, turn the wave on and off as a pulse, be in a material etc. Then it can be more complicated as to E and B are oriented with respect to each other. In general you can get around these issues as a superposition of plane waves as different frequencies and polarizations. With the boundary conditions connecting to some kind of dispersion relation.
Even without boundary conditions, If you have the wave traveling in a material, then you will have dispersion. If the material is anisentropic then the energy flow doesn’t need to be perpendicular to the wavefronts. Your derivation would be a lot more complicated as soon as you have a material with a dielectric constant, especially if the dielectric constant varies with frequency.
Also in a material the group velocity doesn’t need to be equal to the phase velocity.
If you have a wave guide, then you have boundary conditions. Then the velocity will be determined by the boundary conditions, and you will have modes.
So I think your derivation is ok in vacuum and gives a reasonable result in that case,
Also, Yes a standing wave is a good mechanical analog.  The superimposed  traveling waves have a phase velocity that is faster than the standing wave. if the superimposed waves have a slightly different frequency you can see a traveling wave at the difference frequency ( beat wave ) that will be moving at a slower velocity than the superimposed waves.
Not a great explanation, but I guess what I am saying is that with these kind of derivations it is easy to lose sight of when they are applicable in real situations.
A: 
First question: for mechanical waves, is there an analog to this phenomenon?

Yes. This is a called standing wave, regardless of whether it is electromagnetic or mechanical in nature.

Second question: Trying to contradict the above derivation

The issue with the derivation that I see is in trying to use the relativistic mass and relativistic momentum expressions for the electromagnetic field. Leaving aside the issue that modern scientists typically reject the concept of relativistic mass, the formulation you described really only applies for small classical particles with definite velocities. It is fundamentally difficult to assign a velocity to a field, and they are spatially extended rather than being small.
Furthermore, your $\mathcal{E}$ does not have units of energy but units of energy density. Typically in relativity quantities and their densities have different transformation properties and you cannot treat them equivalently.
In this case, the quantity that you want to use is the electromagnetic stress energy tensor: $$ T^{\alpha\beta} = \begin{pmatrix}
\varepsilon_{0} E^2/2 + B^2/2\mu_0 & S_x/c & S_y/c & S_z/c \\
S_x/c & -\sigma_{xx} & -\sigma_{xy} & -\sigma_{xz} \\
S_y/c & -\sigma_{yx} & -\sigma_{yy} & -\sigma_{yz} \\
S_z/c & -\sigma_{zx} & -\sigma_{zy} & -\sigma_{zz}
\end{pmatrix}\,$$ This tensor is always null (traceless ${T^\mu}_\mu=0$), so the EM field is massless. This holds even for a standing wave.
So the fields being null is not contingent on the propagation velocity being c. In standing waves and in the near field we often have scenarios where E is not orthogonal to B and where there is no clear velocity for the wave, but even so the field is still null.
