Why isn't there a centre of charge?
There is no "center of charge" that simplifies calculations completely because there is no center gravity that does this, either.
Two rigid bodies feel gravity that is quite similar to that of point masses, but not exactly the same. Unless the objects are perfect spheres, they feel tidal forces, which depend in second-order and higher derivatives of the potential.
We can only calculate gravitational interactions as points at the center of mass if we want to ignore these second-order effects. On a day-to-day basis, using gravity for things like playing baseball with your nephew, that's fine. The tidal forces are usually very small in applications that we encounter because the phenomena are small compared to the characteristic length scales involved (e.g. a baseball's trajectory is small compared to the radius of Earth, Earth's diameter is fairly small compared to the Earth-Sun distance, etc.)
We can indeed make the same first-order approximation in electrostatics that we do in Newtonian gravity. In practice, a slightly different procedure is normally used, called the multipole expansion. First, a reference point is fixed. The total charge, called the monopole moment, is independent of this reference point, so if you want your first-order approximation to be good, it is wise to choose a reference point near the center of your charge distribution. Higher-order effects are then calculated relative to the reference point by computing dipole, quadrupole, and higher moments of the charge distribution. These higher orders are more likely to come into play in everyday life in electrostatics than in gravity because the size of the charge distributions is similar to the separation between charged bodies (e.g. two balloons you rubbed on your head and are playing with are not very far separated from each other, compared to their diameters).
You might also be interested in the hard sci-fi novel "Incandescence", in which a species of beings living inside a large asteroid orbiting a black hole observe tidal effects to discover general relativity without ever seeing the outside world. It's an interesting demonstrations of what higher-order gravity effects look like beyond the center-of-mass approximation.
The author, Greg Egan, has a web page explaining the tidal effects described in his novel here.
In the same way that bodies that are far away from you can be approximated as point masses, charge distributions that are sufficiently far away can be approximated as point charges, with total charge $Q$, and then you would, indeed, compute something like the center of charge.
This is the first part of the so-called multipole expansion: You begin by approximating a charge distribution $\varrho(x)$ by a point charge of charge $Q = \int dx \varrho(x)$. The next step would be to look at the dipole moment, $p = \int dx x\varrho(x)$, followed by the Quadrupole moment etc
We don't normally talk about the center of charge of distributions because most useful charge distributions are neutral, with total charge $Q=0$, and therefore it doesn't even make sense to define $$ \mathbf r_\mathrm{cc}=\frac1Q \int \mathbf r\rho(\mathbf r)\mathrm d\mathbf r. $$ This means that charged and neutral systems behave very differently in this regard.
For neutral systems, the equivalent quantity of interest is the electric dipole moment, $$\mathbf d=\int \mathbf r\rho(\mathbf r)\mathrm d\mathbf r,$$ which has the property that the leading term in the electrostatic potential at points $\mathbf r$ far away from the charge distribution is given by $$\varphi(r) = \frac{\mathbf d\cdot\mathbf r}{r^3} + O(1/r^3).$$ As opposed to the centre of charge, we do talk about electric dipole moments, all the time.
For a charged system, the dipole moment still plays a role, but it is secondary: far away from the distribution, the electrostatic potential can be better described in the form $$\varphi(r) = \frac{Q}{r}+\frac{\mathbf d\cdot\mathbf r}{r^3} + \frac{p_2(x,y,z)}{r^5} + O(1/r^4), \tag{$*$}$$ where $p_2$ is a homogeneous polynomial of degree 2; these are the first three terms in the multipole expansion of that potential. As you can see, the dipole moment is now relegated to the subleading term, with the leading term given by the charge.
Moreover, for charged systems, the dipole moment becomes less useful as a quantity, because it becomes dependent on the position of the origin; thus, if you displace your coordinate system by $\mathbf r_0$, the dipole moment changes as $\mathbf d\mapsto \mathbf d-Q\mathbf r_0$. This means, therefore, that the centre of charge has the very specific property that
the centre of charge of a charge distribution with nonzero total charge is the position where the distribution's electric dipole moment vanishes.
In terms of the multipole expansion $(*)$, this is important, because the subleading term (which goes as $\sim 1/r^2$) vanishes, so you get a monopole contribution (which goes as $\sim 1/r$) corrected by a quadrupole term which goes as $\sim 1/r^3$, one order of magnitude than the dipole correction.