There is no fallacy, you're just not being particularly careful. You need to include both the electric and magnetic forces of the right magnitude and a covariant result drops out.
(Of course historically it went the other way around: people noticed that frame changes were messed up unless the transformation laws were different and this led to the development of special relativity.)
For simplicity let both beams be very thin and have equal uniform charge density $\rho$ in the rest frame and suppose they run exactly parallel separated by a distance $l$. Let the velocity of the beams be $v$ in the lab frame.
Rest frame:
Taking the usual gaussian pillbox gives the electric field of one beam at the location of the other as
$$ \vec{E}_\text{rest} = \frac{\rho}{2\pi\epsilon_0 l} \hat{r}, $$
where $\hat{r}$ is the unit vector directed away from the source beam. Thus the force on a single particle in the second beam (charge $q$) is:
$$ \vec{F}_\text{rest} = \frac{\mathrm{d}p}{\mathrm{d}t_\text{rest}} = \frac{\rho q}{2\pi\epsilon_0 l} \hat{r}. $$
Lab frame:
The charge density of the beam is enhanced by the relativistic $\gamma=1/\sqrt{1-v^2/c^2}$ factor. Thus the electric field is:
$$ \vec{E}_\text{lab} = \frac{\gamma \rho}{2\pi\epsilon_0 l} \hat{r}. $$
There is also a magnetic field of magnitude
$$ B = \frac{\mu_0 \gamma\rho v}{2\pi l} $$
and directed so as to produce an attractive force. Plugging these in the Lorentz force formula
$$ \vec{F}_\text{lab} = q (\vec{E}_\text{lab} + \vec{v}\times\vec{B}) = q\left( \frac{\gamma \rho}{2\pi\epsilon_0 l} - \frac{\mu_0 \gamma\rho v^2}{2\pi l}\right) \hat{r} = \frac{\rho q}{2\pi\epsilon_0 l}\gamma\left(1 - \epsilon_0\mu_0 v^2\right) \hat{r}. $$
Using $\epsilon_0 \mu_0 = c^{-2}$ this reduces to $\vec{F}_\text{lab} = \gamma^{-1} \vec{F}_\text{rest}$ which, on noting the relativistic time dilation $\mathrm{d}t_\text{lab} = \gamma \mathrm{d}t_\text{rest}$, is exactly right! Note that I've used the fact that the force is orthogonal to the velocity implicity when writing the Lorentz transformation law for the force. You can prove the covariance for general motions using the covariant formulation of EM.
Lesson:
Relativity and electromagnetism go together like hand and glove!
As far as your comment goes, you mean there is an absolute symmetry between the 2 wires. Maybe, but one thing I must tell you that when you are considering the electrons in WIRE 1, the relativistic effects will be as follows:
- The electrons in their reference frame will consider the protons IN WIRE 2 to be in motion.
- Then due to relativistic length contraction, the electrons will observe that WIRE 2 has a higher positive charge density. So the electrons will face more Coulombic attraction from the WIRE 2 than repulsion from the electrons in that wire.
- Most importantly, where I think you are going wrong, the electrons in WIRE 1 will NEVER see the protons in WIRE 1 to be MORE as you say. That is, electrons in WIRE 1 will see that WIRE 2 has a higher positive charge density than WIRE 1 always due to special relativity and nothing else. On the other hand, in a similar fashion, electrons in WIRE 2 will see that WIRE 1 has a higher positive charge density than WIRE 2 always due to special relativity and nothing else.
Hope your doubt has been resolved.
Best Answer
I'm only beginning to study electromagnetism, so please jump on any mistakes I may have made.
Coulomb's law only holds for electrostatic situations, so applying Coulomb's law in the observer's frame of reference is invalid, since the charges are moving from the observer's perspective. The electric and magnetic field of dynamic systems is instead given by the four Maxwell Equations. Even for a simple scenario for this one, the math can get very complicated very quickly, but I'll try to explain the relevant physics.
As the particle moves through space, the electric and magnetic fields in space change as a function of time. Maxwell's equations tell us that changing magnetic fields contribute to the electric field and changing electric fields contribute to the magnetic field. This link between the two fields makes sense from the sense of special relativity since the electric and magnetic fields are relativistically the same field.
After doing all the math (which involves a couple of second order partial differential equations), you will be able to calculate the electric and magnetic forces on each electron and sum them to find that the force is indeed equivalent to the simple Coulombic force from the electron's frame of reference.