Under a proper Lorentz transformation, the Lorentz force 3-vector and 4-vector is not frame independent. Suppose we move to a new reference frame with velocity $\mathbf{V}_{o}$, the electric and magnetic field (in cgs units now) 3-vectors transform as:
$$
\begin{align}
\mathbf{E}' & = \gamma \left( \mathbf{E} + \frac{ \mathbf{V}_{o} }{ c } \times \mathbf{B} \right) - \frac{ \gamma^{2} }{ \gamma + 1 } \frac{ \mathbf{V}_{o} }{ c } \left( \frac{ \mathbf{V}_{o} }{ c } \cdot \mathbf{E} \right) \tag{0a} \\
\mathbf{B}' & = \gamma \left( \mathbf{B} - \frac{ \mathbf{V}_{o} }{ c } \times \mathbf{E} \right) - \frac{ \gamma^{2} }{ \gamma + 1 } \frac{ \mathbf{V}_{o} }{ c } \left( \frac{ \mathbf{V}_{o} }{ c } \cdot \mathbf{B} \right) \tag{0b}
\end{align}
$$
while the 3-vector velocity transforms according to the addition of velocities and $\gamma$ is the Lorentz factor. One can see that Equations 2a and 2b reduce to 0a and 0b in the limit $V_{o} \ll c$ because $\gamma \rightarrow 1$ and the 2nd term in Equations 2a and 2b are 2nd order in $\tfrac{ V_{o} }{ c }$. The $\tfrac{ \mathbf{V}_{o} }{ c } \times \mathbf{E}$ drops out when you convert back to SI units because $B \rightarrow c \ B$ thus there is a factor of $\tfrac{ V_{o} }{ c^{2} }$ in that term.
Now, say I change to a frame of reference that is moving with the negative ions - how does that affect how the electric field should be considered in the analysis?
Electric currents and electromagnetic fields are not Lorentz invariants, so it should not be surprising that the current density (a 3-vector in nonrelativistic cases and a 4-vector in relativistic) changes in relativistic cases.
In the nonrelativistic case, you can think of electric currents as being the relative charge density flux between different species. That is, the nonrelativistic current density can be defined as:
$$
\mathbf{j} = \sum_{s} q_{s} \ n_{s} \ \mathbf{v}_{s} \tag{1}
$$
where $q_{s}$ is the charge [e.g., C] of species $sS (including sign), $n_{s}$ is the number density [e.g., $cm^{-3}$] of species $sS, and $\mathbf{v}_{s}$ is the 3-vector velocity [e.g., km/s] of species $sS in the reference frame of interest.
If you transform to a frame where $\mathbf{v}_{-i} = 0$, that is perfectly okay, but note that $\mathbf{v}_{+i}$ will change to account for the loss of the contribution due to $\mathbf{v}_{-i}$. In the absence of nuclear/quantum effects (e.g., pair production), the number density and charges are all conserved so a Galilean transformation only changes $\mathbf{v}_{s}$. Since it's a linear operation (i.e., subtract all $\mathbf{v}_{s}$ by $\mathbf{v}_{-i}$), the change is only in the bulk flow velocities of each species.
Now does it remain constant? Let's try a simple thought experiment. Suppose you have two, oppositely signed charge carriers drifting along $\hat{\mathbf{x}}$ relative to each other. Further suppose they have equal but opposite charges and equal number densities. Does there exist a reference frame where the net current is zero? If both species drift in the same direction then $\mathbf{j} \rightarrow 0$ when the magnitudes of their drift velocities match, i.e., there is no relative drift between the two species. If the two species have different charge states (e.g., alpha-particles and electrons), then they need not drift together to have a zero net current so long as they drift the same direction at the correct ratio of speeds.
So, do I now need to have two different E fields in the new frame, for the positive/negative ions?
No, as I said above the Lorentz force is not a Lorentz invariant. That is, the forces acting on each charge species will change in different reference frames.
Or, is it the apparent charges of the particles that have changed?
No, total charge is a conserved scalar quantity under a Lorentz transformation. Charge density, however, can change under a Lorentz transformation. Though the charge density cannot change arbitrarily, as it is part of the 4-vector with the 3-vector current density to make up what's sometimes called the 4-current. Under Galilean transformations, the charge density is conserved.
So in the limit as $\tfrac{ V_{o} }{ c } \rightarrow$ very small (i.e., Galilean transformations), velocities, and the electric and magnetic fields transform as:
$$
\begin{align}
\mathbf{E}' & = \mathbf{E} + \mathbf{V}_{o} \times \mathbf{B} \tag{2a} \\
\mathbf{B}' & = \mathbf{B} \tag{2b} \\
\mathbf{v}' & = \mathbf{v} - \mathbf{V}_{o} \tag{2c}
\end{align}
$$
Note that Equation 2a shows that there can exist reference frames with no electric fields, if the only electric fields present are quasi-static.
