Let us consider a lattice of static positive ions, through which electrons flow. In the laboratory reference frame, we will have equal charge densities and different velocities, thus their 4-currents would be:
$j_+^\mu=(\rho_+,0)$
$j_-^\mu=(\rho_-,\rho_-\mathbf{v}_-)=(-\rho_+,-\rho_+\mathbf{v}_-)$
and the total current, which creates an electromagnetic field, is their sum:
$j_\Sigma^\mu=j_+^\mu+j_-^\mu=(0,-\rho_+\mathbf{v}_-)$
So far, it is as we expected. Now, let us switch to the moving reference frame, having the velocity $\mathbf{v}_-$, so that electrons would appear static. That is done by the Lorentz transformations for 4-vector of current, and the result is ($\gamma=1/\sqrt{1-v_-^2}$ is the usual abbreviation):
$j_+^{\,\prime\mu}=(\gamma\rho_+,-\gamma\rho_+\mathbf{v}_-)$
$j_-^{\,\prime\mu}=(\gamma\rho_--\gamma\rho_-v_-^2,\gamma\rho_-\mathbf{v}_--\gamma\rho_-\mathbf{v}_-)=(-\rho_+/\gamma,0)$
(zero current of electrons, as we expected)
$j_\Sigma^{\,\prime\mu}=j_+^{\,\prime\mu}+j_-^{\,\prime\mu}=(\rho_+v_-^2\gamma,-\gamma\rho_+\mathbf{v}_-)$
This is exactly the same as we would get if applied Lorentz transformations immediately to $j_\Sigma^\mu$. No balance of the charge in a moving frame is left.
Now we can consider some other conducting system, say, a cell with electrolyte, adjusted in such a way that positive and negative ions in it have the same density and opposite velocity vectors. Now we repeat the same calculations and get:
$j_+^\mu=(\rho_+,\rho_+\mathbf{v}_+)=(\rho_+,-\rho_+\mathbf{v}_-)$
$j_-^\mu=(\rho_-,\rho_-\mathbf{v}_-)=(-\rho_+,-\rho_+\mathbf{v}_-)$
$j_\Sigma^\mu=j_+^\mu+j_-^\mu=(0,-2\rho_+\mathbf{v}_-)$
$j_+^{\,\prime\mu}=(\gamma\rho_++\gamma\rho_+v_-^2,-\gamma\rho_+\mathbf{v}_--\gamma\rho_+\mathbf{v}_-)=(\gamma\rho_+(1+v_-^2),-2\gamma\rho_+\mathbf{v}_-)$
$j_-^{\,\prime\mu}=(\gamma\rho_--\gamma\rho_-v_-^2,\gamma\rho_-\mathbf{v}_--\gamma\rho_-\mathbf{v}_-)=(-\rho_+/\gamma,0)$
$j_\Sigma^{\,\prime\mu}=j_+^{\,\prime\mu}+j_-^{\,\prime\mu}=(2\rho_+v_-^2\gamma,-2\gamma\rho_+\mathbf{v}_-)$
We got the same result! (With the factor of 2 for both frames.) That means, it does not depend on the way we construct the current. The fact holds true that if charges are balanced in some chosen reference frame, then they become unbalanced in other reference frames.
Now, why are charges balanced in the laboratory reference frame, what it the physical cause for this? The moving charges are ultimately supplied to the wire by some current source. That source produces the same amount of current carriers on one pole, as it absorbs on the other pole, and thus keeps the wire uncharged as a whole. Would it be different, that would take much energy to keep the wire in a charged state, because charged conductor has an electrostatic energy $q^2/2C$. That current source "calculates" the energy and balance of chanrge in its own reference frame. That's it: would the wire and the source move, they would keep some other balance, zero in a different reference frame.
Update: Let's consider contraction and stretch of individual charged particles. We follow the Wikipedia article. Figures are taken from it.
In the lab frame:
Negative charges are "the lattice" and they make a chain with positions $x_k=ka$, where $a$ is the lattice constant. Then their world-lines are given with equations
$x_{-,k}^\mu(t)=(t,ka\mathbf{e}_x)=(t,ka,0,0)$
Note that the lattice constant is not arbitrary here, and is fixed by the nature of the lattice (for example, if the lattice is made of Fe ions, $a$=0,286645 nm).
Positive charges are "the flowing charges", and they make a chain with the same spacing, though moving. The positions $x_k=ka$ would be only their initial positions, and the world-lines are
$x_{+,k}^\mu(t)=(t,(ka+vt)\mathbf{e}_x)=(t,ka+vt,0,0)$
The requirement $x_{+,k}^x(0)=x_k$ is imposed in order for the wire to have zero charge as a whole.
In the moving frame:
To get this picture, we should take the world-lines of charges in the lab frame, and apply Lorentz transformations to them. For the negative ("lattice") charges, we have
$x_{-,k}'^\mu(t')=(t',\gamma ka-vt'-\gamma kav^2,0,0)=(t',ka/\gamma-vt',0,0)$
for the positive ("flow") charges, respectively,
$x_{+,k}'^\mu(t')=(t',\gamma ka,0,0)$
(please check these calculations yourself).
That shows that the negative ("lattice") charges are moving backwards with the speed $v$, and their distance is contracted ($a/\gamma<a$). At the same time, the positive ("flow") charges are static in the moving frame, and their distance is stretched ($\gamma a>a$). The factor $\gamma$ is a number greater than 1.
Finally, the negative ("lattice") charges are contracted with respect to the positive ("flow") charges ($a/\gamma<\gamma a$) in the frame of the positive ("flow") charges. And the positive ("flow") charges are not contracted with respect to the negative ("lattice") charges in the frame of the negative ("lattice") charges, they do have the same spacing ($a=a$).
