Hello, Zeus! I'm so glad to come across your problem. Nice and attractive!
It is essential in special relativity that, a free electron can neither absorb nor emit photons. Hints are given in details below.
1. Absorbing of photon is forbidden.
If a free electron could absorb a photon, then, according to conservation of energy and momentum,
$$
\hbar \omega +mc^2 = \sqrt{p^2c^2 + m^2c^4} (Eq1)
$$
$$
\hbar k = p (Eq2)
$$
where $\omega$ and $k$ are the frequency and wavenumber of the photon, respectively, $m$ the electron's rest mass, $p$ the momentum of the electron after absorbing the photon. Eq2 leads to
$$
p^2=\hbar^2 k^2 = \hbar^2 \frac{\omega^2}{c^2} (Eq3)
$$
Insert Eq3 into Eq2, and square of the left side is
$$
(\hbar \omega +mc^2 )^2 =\hbar^2 \omega^2 + m^2c^4+2mc^2\hbar \omega
$$
when square of the right side is
$$
\hbar^2\omega^2+m^2c^4
$$
So, if Eq1 holds, $\hbar \omega=0$. There is no photon carrying vanishing energy. Hence, absorbing of a photon by a free electron is forbidden.
2. Emitting of a photon is Forbidden
Suppose the intial (before photon emitting) and final (after photon emitting) 4-momentum of the electron are separately $p_\mu$, $p'_\mu $:
$$
p_\mu=(0,0,0,imc), \qquad p'_\mu = \left[p, i\frac{mc}{\sqrt{1-\frac{v^2}{c^2}}} \right] (Eq4)
$$
According to conservation of energy and momentum:
$$
p_\mu=q_\mu+p'_\mu (Eq5)
(p_\mu - p'_\mu)^2 =q_\mu^2 (Eq6)
$$
where $q_\mu$ refers to the 4-momentum of the photon. From Eq5 and Eq6, we have
$$
(p_\mu - p'_\mu)^2 = p_\mu^2 -2p_\mu p'_\mu+p'_\mu^2
=-m^2c^2- 2p_\mu p'_\mu -\frac{mc^2}{1-\frac{v^2}{c^2}} (Eq7)
$$
Recall that
$$
p_\mu p'_\mu = -\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} (Eq8)
$$
Insert Eq8 into Eq7, we have
$$
p_\mu p'_\mu = -m^2 c^2 \left[ 1-\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \right] (Eq9)
$$
Yet $q_\mu^2=0$, so, $p_\mu=p'_\mu$, to ensure Eq5 and Eq6 hold. Thus the state of motion of the electron is not disturbed, with no momentum transported to the photon. Hence, emitting of a photon by the free electron is forbidden.