If we are using [eV] as a unit for energy, what, then, should the unit for mass and distance be? This seems obvious but it is confusing since I know that in MKS (i.e. SI) system, we use Joule with Meter with KG.
But if we're using electron volt (1.6e-19 J) for energy, should we change the mass and distance units too?
 A: It depends on what you mean by "should."  You can, if you'd like, quote all of your energies in $\mathrm{eV}$, all of your masses in $\mathrm{kg}$, and all of your distances in $\mathrm{m}$ without any inconsistency.
However, you're probably referring to the fact that joules are naturally suited to kilograms, meters, and seconds in the sense that
$$
  \mathrm J = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}
$$
without any weird dimensionless prefactors on either side of the equation, and you want to know what mass and length units, lets call them $\mathrm M_\mathrm{eV}$ and $\mathrm L_\mathrm{eV}$ give you a relationship similar to the relationship for $\mathrm{J}$;
\begin{align}
  \mathrm{eV} = \frac{\mathrm M_\mathrm{eV}\cdot \mathrm L_\mathrm{eV}^2}{s^2}\qquad  ?\tag{$\star$}
\end{align}
If so, then note that there isn't a unique pair that does this.  As you noted, there is a dimensionless constant $a$ relating Joules and electron volts;
$$
  a\cdot \mathrm{eV} = \mathrm J
$$
This means that we can use the first relation above to write
$$
  a \cdot\mathrm{eV} = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2}
$$
Now we notice that both of the following pairs of choices will lead to an expression like $(\star)$: for choice 1 take
$$
  \mathrm M_\mathrm{ev} = a^{-1}\cdot \mathrm{kg}, \qquad \mathrm L_\mathrm{eV} = \mathrm m
$$
and for choice 2 take
$$
  \mathrm M_\mathrm{ev} = \mathrm{kg}, \qquad \mathrm L_\mathrm{eV} = a^{-1/2}\cdot \mathrm m
$$
