Suppose you start with a linear charge density $\lambda^+$ of positive charges and $-\lambda^-$ of negative charges in the wire, everything at rest.
Case 1: No current, test charge stationary
You assume you have a neutral wire with no current. Therefore $\lambda^- = \lambda^+$. There's no other frame worth considering, since nothing is in motion anyway.
Even if you did go into another frame, any change in charge density will affect electrons and nuclei equally. Thus the wire is neutral in all frames, and test charges are entirely unaffected by it.
Case 2: Nonzero current, test charge moving with electrons
Now suppose you have a wire with a current. Again, the wire is neutral in the lab frame $S$, where the bulk of it is not moving. In this frame, we still must have $\lambda_S^- = \lambda_S^+$, even though the electrons are moving and the nuclei aren't.
If we slip into the rest frame $S'$ of the bulk electron motion, then the spacing between electrons must be different, and in fact it must be larger. Since charge doesn't change when changing frames, we know $\lambda_{S'}^- < \lambda_S^-$. Similarly, the nuclei spacing will be length-contracted, so $\lambda_{S'}^+ > \lambda_S^+$. In this frame, then, $\lambda_{S'}^+ > \lambda_{S'}^-$, so the wire looks positively charged, and any (positive) test charge at rest in this frame $S'$ will be repelled.
As you can check, this is exactly what the Lorentz force law tells you. If the electron bulk motion is in the $-z$-direction, then the current is in the $+z$-direction, and the magnetic field along the $+x$-axis (assuming the wire coincides with the $z$-axis) is in the $+y$-direction. A positive charge with velocity in the $-z$-direction in a magnetic field in the $+y$-direction will experience a force in the direction of $(-\hat{z}) \times (+\hat{y}) = +\hat{x}$, away from the wire.
Case 3: Nonzero current, test charge stationary
Now consider the setup as follows. In $S$, the nuclei and test charge are stationary, but the electrons are moving in the $-z$-direction. Just as before, we can transform into the electrons' rest frame, where we will find that the wire is positively charged. However, we also have that the test charge is moving in the $+z$-direction in $S'$, and that there is a current of positive charges in the $+z$-direction (which we could neglect earlier). Here the full Lorentz force law tells us there is a $qE$ repulsion, and also a $q \vec{v} \times \vec{B}$ attraction, and in fact they perfectly balance in this frame, so there is still no net force.
Summary
The space between electrons expands only if you keep yourself in their rest frame as you accelerate them. The spacing measured by an observer who doesn't accelerate is unchanged, in keeping with the assumption that the wire stays neutral in the lab frame. You can only use the electrostatic Coulomb's law if you are in the frame where the test charge of interest is stationary. If you are in a frame where the charge is still moving, you need the full Lorentz law, using whatever electric and magnetic fields are present in that frame.
But the row of moving electrons is contracted in the lab frame, compared to what the cat sees. You can see that in your screencaps from the video too -- Derek sees 10 electrons per image width, whereas in the rest frame of the electrons there are only about 8½ electrons per image width.
What is potentially confusing is that as far as the electrons themselves are aware (electrons are not "aware" of anything, but never mind), they are not that the same mutual distance when they are moving as when the wire carried no current.
In other words, the row of electrons is not a rigid object. If each pair of neighboring electrons had been separated by a little rigid rod, the electrons would have to come closer together when the current starts flowing. But there are no such rods, and the row of electrons is free to stretch when the current begins to flow, and this stretching is exactly canceled out by the length contraction, such that in the lab it looks like the distance between the moving electrons is the same as the distance between the stationary protons.
What does the cat see? When the wire didn't carry current, the electrons and protons moved backwards together at the same speed (and with the same contracted spacing as the cat sees the protons have during the entire experiment). Then, when the current starts to flow, the electrons in front of the cat begin coming to a halt (with respect to the cat) slightly before those behind it. So from the cat's point of view the row of electrons get significantly stretched.
Meanwhile, Derek will see all the electrons begin to move at the same time. The cat and Derek do not agree whether two electrons changed their velocity at the same time or not -- this is relativity of simultaneity and is mathematically necessary to make length contraction consistent.
Best Answer
Model: Let's simplify the model of a current in a wire, so we can be definite about what we are talking about. Take a wire (in the wire's frame) to have fixed positive charge density $\rho_{+}$ and assume the electrons at rest w.r.t the wire, with electron density $\rho_{-}$.
Introducing a current sets these electrons moving at some speed $v_\rm{drift}$ w.r.t wire, but leaves the positive charge fixed. We ask the following question:
What is the relationship between $\rho_{-}$ (the electron charge density at rest), and the electron density with current?
Answer: The density measured by the observer stationary w.r.t a current carrying wire is not the same as if the charges were stationary. They are related by a Lorentz transformation. Let's write the 4-current of the electrons when at rest, and when moving (with $c=1$): $$J^\mu_\textrm{rest} = (\rho_{-},\vec{0})^\mu,\,\,\,\,\,J^\mu_\textrm{moving} = (\tilde{\rho},\vec{j})^\mu = {\Lambda(v)^\mu}_\nu J^\nu_\rm{rest}$$ where $\Lambda(v)$ is the Lorentz Boost between these two frames. Note in particular that $\boxed{\rho_{-} \neq \tilde{\rho}_{-}}$ because $$J^2_\textrm{rest} = J^2_\textrm{moving}~~\implies ~~ \rho_{-}^2 = \tilde{\rho}^2_{-}-\vec{j}.\vec{j},$$ and $~\vec{j}\neq \vec{0}$.
This means that when you set up your problem, we have two possible scenarios:
$(i)$ $\rho_{-}+\rho_{+} = 0$, that is we ask that the electron density in the electrons rest frame has the same magnitude as the positive charge density in the stationary wire.
$(ii)$ $\tilde{\rho_{-}}+\rho_{+} = 0$, that is we ask that the electron density in the wire's rest frame has the same magnitude as the positive charge density in the stationary wire. This is the situation of zero force on a stationary external charge you talked about in your edit.
So the question you have to ask yourself, is what situation do you want to deal with? It seems that for the "explanation of magnetic force as a consequence of special relativity" you are interested in, one should consider case $(ii)$ as this allows you to see how a test charge, moving parallel to the wire with velocity $v$, experiencing a force due to a pure magnetic force in one frame (wire rest frame) $F = q v\times B$, is the same force experienced by the charge in its rest frame, effected only by the electric force, $F = q E$, in that frame (as in this frame it isn't moving).
I hope this helps. If you need further explanation, don't hesitate to ask.