The metric.
A charged blackhole follows Reissner–Nordström metric, which is:
$$
c^2 d\tau^2 = \left(1 - \frac{r}{r_s} + \frac{r^2}{r_Q^2}\right) c^2 dt^2
- \left(1 - \frac{r}{r_s} + \frac{r^2}{r_Q^2}\right)^{-1}dr^2 - r^2 d\theta^2 - r^2\sin^2\theta d\phi^2
$$
Where:
$$
r_s = \frac{2GM}{c^2},\quad\quad r_Q^2 = \frac{GQ^2}{4\pi\epsilon_0 c^4}
$$
Notice if the charge is zero $Q=0$, then $r_Q = 0$, and we are back to a Schwarzschild metric, the usual metric around blackholes. For every discussion we gonna make, feel free to set $r_Q = 0$, in order to visualize what would happen to a 'normal' blackhole.
Just so we know what we are talking about, I better explain the concept of a metric. It measures the spacetime distance, or, the proper time $d\tau$, between two events very close to each other. An event is a spacetime point $(t,x,y,z)$, or, in spherical coordinates, $(t, r, \theta, \phi)$.
Example, if the events are separated by $dt = t' - t$ and $dr = r' - r$ and so on, then, the spacetime difference is given by above formula. Of course, $r'-r$ must be pretty small for this to work, and the formula is exact for infinitesimal distances.
The horizon.
The problem, is the $dr$ term. Look at it! The term that multiplies $dr^2$ in the metric, is something: $(\cdot)^{-1}$. There's a power -1. In other words, its in the denominator: $1/(\cdot)$. What happens if the denominator is zero? Then it means for an infinitesimal distance $dr$, we have potentially infinite spacetime distance $d\tau$. Do you get it? Yes! You do! We are talking an event horizon here. An horizon, where events are not accessible, as it takes an infinite amount of time to be seen.
So let's do it! Let's make it zero and locate the horizon. In the uncharged case, $Q = 0$, $r_Q = 0$, we have:
$$
1 - \frac{r}{r_s} = 0
\quad\implies\quad
r = r_s
$$
Thus, the event horizon of the blackhole is a spherical surface with radius equal the Schwarzschild radius, $r_s$.
Thankfully for us, for the charged case, it is just a quadratic formula:
$$
1 - \frac{r}{r_s} + \frac{r^2}{r_Q^2} = 0
\quad\implies\quad
r_{\pm} = \frac{1}{2}\left(r_s \pm\sqrt{r_s^2 - 4r_Q^2}\right)
$$
Keep in mind that, the greater the charge $Q$, the greater the length $r_Q$ is. We can analyze the discriminant $\Delta$.
If $r_s > 2r_Q$: Just a little bit of charge. In this case, the discriminant is positive, $\Delta > 0$. There are two horizons. The usual event horizon $r_+$, and an inner horizon, called Cauchy horizon $r_-$.
If $r_s = 2r_Q$: A specific amount of charge. In this case, the discriminant is zero, $\Delta = 0$, and both horizons meet: $r_+ = r_- = \frac{1}{2}r_s$. This blackhole is called an extremal blackhole.
If $r_s < 2r_Q$: Lots of charge. In this case, there are no more horizons, as $\Delta < 0$. There's no real solutions to this quadratic equation. This means, it isn't a blackhole anymore.
The forces.
The ratio of gravitational and electric forces are:
$$
\frac{F_m}{F_q} =
\frac{GMm}{r^2}\frac{4\pi\epsilon_0 r^2}{Qq} =
4\pi G\epsilon_0\frac{M}{Q}\frac{m}{q} =
4\pi G\epsilon_0\left(\frac{r_s c^2}{2G}\sqrt{\frac{G}{4\pi\epsilon_0 c^4 r_q^2}}\right)\frac{m}{q}
$$
Thus:
$$
\frac{F_m}{F_q} =
\frac{1}{2}\frac{m}{q}\frac{r_s}{r_Q}\sqrt{4\pi\epsilon_0 G}
$$
Where $\sqrt{4\pi\epsilon_0 G}\approx 8.61\cdot 10^{-11} C/Kg$. Normally, $m/q\ll 1$, for instance, for an electron, $m/e\approx 5.68\cdot 10^{-12} Kg/C$. And, assuming extremal blackhole, $r_s/r_q = 2$, we then have $F_m/F_q\ll 1$. So, electrostatic forces are much larger than gravitational forces.
Of course, this is a naive calculation. To get the real thing, I'd have to calculate the effective potential of this blackhole as to figure out exactly what are the conditions for like charge to enter the black hole and increase its charge.
Effective potential.
For that we need the equations of motion, and thus we need to calculate the extremum of the action.
$$
S = \int Ld\tau = \int g_{\mu\nu}(x) dx^\mu dx^\nu = \int\left[-A(r)c^r\dot t^2 + A(r)^{-1}\dot r^2 + r^2\left(\dot\theta^2 + \sin^2\theta\dot\phi^2\right)\right] d\tau
$$
Where here it was considered $A(r) = 1 - \frac{r}{r_s} + \frac{r}{r_Q^2}$. Actually, this is not exactly the action (we missed a square root), but, it can be shown that the extremum of the real action coincides with the extremum of this action $S$ that we are working with. So, we are good, if our goal is to find quantities only dependent on the action extremum, such as, the equation of motion. The integrand can be seen as a Lagrangian, thus, we can plug it in Euler-Lagrange equations, and figure out everything we need. The calculations are not hard, but, I shall omit them. After doing all of that, we get the final answer:
$$
\frac{1}{2}\dot r^2 + \frac{1}{2}A(r)\left(\frac{\ell^2}{r^2} + c^2\right) = \frac{1}{2}\frac{E}{c^2}
$$
Where the conserved quantities $\ell$ and $E$ taken from the $\theta$ and $t$ Euler-lagrange equations, are exactly the angular momentum and energy of the system. We can compare the above equation with the classical equations of motion, and thus we can identify the effective potential from the middle term $\frac{1}{2}A(r)\left(\frac{\ell^2}{r^2} + c^2\right)$. Thus, plugging in for $A(r)$, we finally have it:
$$
V_{eff}(r) = \frac{1}{2}\left(1 - \frac{r_s}{r} + \frac{r_Q^2}{r^2}\right)\left(\frac{\ell^2}{r^2} + c^2\right)
$$
Plugging in $r_s$ and $r_Q$, and, expanding, we get:
$$
V_{eff}(r) =
\frac{c^2}{2} + \frac{\ell^2}{2r^2}
- \frac{GM}{r} - \frac{GM\ell^2}{r^3 c^2}
+ \frac{GQ^2}{8\pi\epsilon_0 c^2 r^2} + \frac{GQ^2\ell^2}{8\pi\epsilon_0 r^4 c^4}
$$
Looking at this, we can identify several terms that would show up in a newtonian theory, and several other new terms, which are there because this is general relativity. Furthermore, this is not exact, as we still would have to introduce the electrostatic potential in an ad hoc way, because we didn't put the electrostatic contribution directly into the action (it would further complicate the action significantly as I would have to include 4-potential terms in the action like $q A_\mu\dot x^\mu$ and, I am lazy, okay?!).
From here, all we do, is to figure out the maximum and minima of the potential, so as to figure out what is the energy barrier a particle with angular momentum $\ell$ would need to overcome, in order to get inside the charged black hole. And thus, if we set the particle with greater energy, it gets through. In order words, we should compute the derivative and set to zero, that is, $V'(r) = 0$. I shall cleverly stop here because I am lazy, because finding the potential maximum/minimum here would involve solving a third degree polynomial.
However, you could, if you want, plot $V_{eff}$, to see how it behaves. Or, if you want to solve it, go ahead.
Let's now answer your questions.
Is such an object possible to create?
Oh well. I don't see any GR objections of creating such an object. So, if you are asking it is possible to create one? Yes. Probably. Maybe.
However, if you are asking: an object like that exists out there? Unlikely. Very unlikely. Very very unlikely. No, it doesn't! Without external interference, nature tends to neutralize charge pretty quickly.
If such an object were to exist, would the electrostatic force make additions to the mass of the hole impossible?
Probably not. It would be difficult though. Like we've done, $F_m/F_q\ll 1$, so electrostatic forces dominate. Like I said before, without external interference, nature tends to neutralize charge pretty quickly.
Even though I didn't prove this, I suspect a particle with sufficient energy would be able to overcome the electrostatic potential barrier and get across. Thus, not impossible.
[...] considering that the mass beyond the event horizon is gone forever, meaning the electrostatic force is irrelevant beyond the event horizon.
You can assume that, yes. But, in reality, it does not happen. Black-holes do evaporate due to Hawking radiation.
Even if such a charge of the black hole were to blow the mass apart normally, the event horizon makes it impossible.
Even if we assume black holes do not evaporate....
What horizon? Its gone! Like we've shown, if there's too much charge such that $2r_Q > r_s$, such an object would no longer have an horizon. So, charge is free to fleeeeeeeee.
