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.
Not all problems work out nicely in 3 + 1 dimensions. Consider two 4-currents in the (unprimed) wire rest frame:
$$ j^+_{\mu} = (\rho^+, 0, 0, 0) $$
$$ j^-_{\mu} = (\rho^-, v_d\rho^-, 0, 0) $$
With the charge densities satisfying $\rho^+ = -\rho^-$ (the wire is neutral) and the drift velocity is leftward, $|v_d| = -v_d$, the total four current density is:
$$ j_{\mu} = (0, v_d\rho, 0, 0) $$
(note: many complain the drift velocity is too small to invoke $\gamma$, which is true, but if you do the Lorentz transform parallel to the current you will see it is the relativity of simultaneity that dominates the change in charge density).
The moral of that parenthetical story is: when in doubt, Lorentz transform. So that's what we need to do here. With ($c=1$) and the test charge's (primed) frame moving along $+y$:
$$ u_{\mu} = \gamma(1, 0, v, 0) $$
you get:
$$ \Lambda_{\mu\nu} = \left [\begin{array}{cccc}
\gamma & 0 &-v\gamma & 0 \\
0 & 1 & 0 & 0 \\
-v\gamma & 0 & \gamma & 0 \\
0 & 0 & 0 & 1
\end{array}\right ]$$
$$ j'_{\mu}= \Lambda_{\mu\nu}j_{\nu} $$
$$ j'_{\mu}= (0, v_d\rho, 0, 0) = j_{\mu} $$
So the force in the primed frame is caused by a magnetic field, no 3 + 1 magic here.
Of course, this is wholly unsatisfying compared with the original problem, which works out in spite on being not at all manifestly covariant.
It also raises the question of "where is the moving charge?". If you're moving above the wire, then the transformation of a magnetic field is:
$$ B'_{\parallel} = B'_{\parallel}$$
That is fine, but if you're moving towards the wire (in the plane of the current and charge):
$$ B'_{\perp} = \gamma\big(B_{\perp} - \vec v \times \vec E\big) =\gamma B_{\perp}$$
as $\vec E = 0$.
So now we've picked up a $\gamma$: the primed field is stronger (and not circular symmetric about the wire). That is not evident from the simple transformation of $j_{\mu}$.
I think the resolution here is to go to Jefimenko's Eq for a magnetic field generated by currents and charges on the past light cone:
$$\vec B'(\vec r', t') = -\frac{\mu_0}{4\pi}\int{\Big[
\frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^3} \times \vec J'(\vec r'', t'_r) +
\frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^2} \times \frac 1 {c}
\frac{\vec \partial J'(\vec r'', t_r)}{\partial t'}
\Big]dV''}$$
Since the current is constant:
$$\vec B'(\vec r', t') = -\frac{\mu_0}{4\pi}\int{\Big[
\frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^3} \times \vec J'(\vec r'', t'_r)\big]dV''}$$
(Note the double primed coordinate here is the integration variable, not
any reference frame).
Computing this a bit of a project. In the primed frame you have a current density looking like:
$$\vec j'(x', y', z', t') = v_d Q\delta(y'-vt')\delta(z')\hat x' $$ where Q is the limit of $\rho$ in an infinitely thin wire. So this is a line of moving current.
Then fix ${\vec r'}$ to be the position of the charge, in the primed frame:
$$ \vec r' = (x', y', z') = (0, 0, 0) $$
At this point, it is best fix $t' \ne 0$ so you have some distance from the wire.
Then compute the retard time $t'_r$. The $\delta(y'-vt')$ will filter over the past light cone along the moving wire, giving some kind of conic section, but it should all work out, as it always does. Have at it. $\gamma$ is in there, somewhere.
Best Answer
So imagine two point, positive charges at rest, in a frame S: the interaction is purely Coulombic, and the force on one another has magnitude $\frac{q}{4\pi\epsilon_0r^2}$, $r$ is the distance between them. Now consider the events from a frame, which I denote as $\bar S$, moving along the x-axis at velocity v: the charges are moving according to this frame. Now, the transformation for fields in special relativity is as follows: suppose one has a frame $S$ with electric and magnetic fields denoted by $(E_x, E_y, E_z)$ and $(B_x, B_y, B_z)$ and another frame $\bar S$ moving at speed v, along the x-axis, relative to S, where the fields are given by $(\bar{E_x},\bar{E_y},\bar{E_z})$ and $(\bar{B_x},\bar{B_y},\bar{B_z})$. Then,
$\bar{E_x} = E_x, \bar{B_x}=B_x$
$\bar{E_y} = \gamma(E_y-vB_z), \bar{B_z}=\gamma(B_z-\frac{v}{c^2}E_y)$
$\bar{E_z} = \gamma(E_z+vB_y), \bar{B_y}=\gamma(B_y+\frac{v}{c^2}E_z)$
This derivation can be found in textbooks like Griffiths' Introduction to Electrodynamics or Feynman's Lectures. In our case, from $S$ there is only an electric field, which will have three components, and a magnetic field which is zero. However, when looking from $\bar S$, there are electric as well as magnetic fields due to the transformation equation I have stated. So yeah, the Coulombic field does change, and we also have a magnetic field here!
Edit: Okay, I am so sorry, I understood your question now, and I'm removing the previous edit. As someone else has remarked here, there is a transformation law for the perpendicular force acting on the particle, which goes as $\bar F = \frac{F}{\gamma}.$ Here, what's important is that from one of the frames, the particle must be at rest, and F is the force on the particle as measured from this frame and $\bar F$ in the other frame. Now for your case, imagine two charges moving at some velocity $\bar u$ along the x-axis relative to a frame $\bar S$, one above the other. $\bar S$ itself is moving at velocity v along the x-axis relative to another frame $S$. What's special about S is that the charges are at rest from this frame(This translates to $\bar u = -v$, but I'll leave you to worry about how that happens!). So, the force between the charges is purely electrostatic, and I will call the direction of the force along the y-axis, and $F=\frac{q^2}{4\pi\epsilon_0y^2}$, say, on the upper charge. Now, from our rule, $\bar F = \frac{F}{\gamma}$, we find that in the frame $\bar S$, the force is reduced. However, we know that by applying relativity to a single charge, the electric field from $\bar S$ should be $\gamma E$, where $E$ is the field from $S$. so we expect the force to also increase by $\gamma$, which does not happen. This hints at the presence of another frame-dependent force acting opposite to the electric field in this frame, and it's of course what we call a 'magnetic force!' Hope this helped!