Irreducible mass of a charged black hole merger The irreducible mass $ \rm m_{irr} $ of a black hole with charge $ \rm Q $ is (in natural units)
$$ \rm m_{irr} = \frac{r_{+}}{2} = \frac{M}{2} + \sqrt{\frac{M^2}{4} - \frac{Q^2}{4}} $$
which is less than the mass equivalent of the total energy $ \rm M $ (as measured from infinity), since $ \rm M $ contains not only the mass of the material but also the mass equivalent of the electromagnetic field energy.
Suppose we merge (neglecting gravitational waves) two charged black holes with the same mass and the opposite charge (so that the resulting net charge should be zero), would the resulting Schwarzschild black hole with $ \rm Q=0 $ have the total mass equivalent of $ \rm 2M $, or rather $ \rm 2 m_{irr} $?
In the reference Horizon Mass Theorem it says

so I would suspect the resulting mass equivalent to bei $ \rm 2 m_{irr} $, but I've also heard some arguments favoring $ \rm 2 M $
 A: This paper studies this question: https://arxiv.org/pdf/1311.6483.pdf
Note that the in-fall process for the configuration that you described results in both gravitational and electromagnetic radiation, both of which carry some energy away from the final black hole.  The precise amount of each depends on the ratio $Q/M$.  It looks like it's relatively small in all cases, so you'll end up with a black hole close to $2M$ in your notation. (The paper starts with each black hole having mass $M/2$ and so gets a final mass close to $M$ in their notation.)
EDIT
Here's some additional explanation following revision to the original question and the ensuing comments.
General relativity is a non-linear theory, so you cannot just take a linear combination of solutions like you can with, for example, the Maxwell equations.  So let's consider to two limiting cases first:


*

*The black holes are "very close" together.

*The black holes are "very far" apart.


Quantifying "very far" and "very close" is beyond what I'll do here, but it should be some multiple of $M$.
When they are very close together, then the expectation should be that they are already approximately Schwarzschild with mass $2M$ (and no charge).  The definition of "irreducible mass" provided by the OP should be applied to the binary (if at all) rather than separately to the individual black holes with their net charge.  But since the net charge is 0 in this case presented, the irreducible mass is also $2M$.  I see no sensible way to apply the definition of irreducible mass separately to the two black holes in this case for a variety of reasons: 


*

*The energy of a charge "lowered" into one of the black holes is path-dependent, so, at best, the expression would need to be more complicated.  

*The whole point of the definition is to separate the energy attributable to the EM field from everything else.  In this case, the EM field is due to both black holes, so it's not immediately clear how such a definition could be made for one of the black holes. (Maybe for equal magnitude charge it's possible, but I think the nonlinearities may make it hard to separate outside of that case.)

*To whatever extent you can compute separate contributions to the EM field from each BH, they will give partially cancelling contributions to the EM field.  Since the irreducible mass is effectively trying to capture the energy in the EM field, whatever you did here would have to account for the fact that the EM field contains much less than the sum of the individual fields due to cancelling effects.  Naively applying the definition is "more like" assuming that the field is doubled.


When they are very far apart, each experiences minimal curvature due to the other.  In this case, you can approximately take the linear combination of solutions as initial data.  For the same reason, it probably makes sense to talk about their individual "irreducible masses" (as defined by the OP).  If you let them fall toward each other, however, a few things will happen:


*

*They will radiate gravitational radiation up to and past the time of merger, which will carry away the non-spherically-symmetric features of the spacetime structure.

*They will radiate electromagnetic energy, which will carry away the non-spherically-symmetric portions of the EM field.

*Some - as it turns out most - of the radiation will fall into the BHs and contribute to the mass of the steady-state, merged BH.


So in the "very far" case, you'll still end up with a merged black hole with mass that is approximately $2M$.  As noted in the paper linked and in the comments, it is not necessarily obvious a priori how much of the energy would have been carried away, but the numerical results suggest that it's a small fraction.
Outside of these extreme cases, you'll need a computer to try to answer.  You probably even need a computer to generate the initial data prior to the in-fall process.
Now I guess, perhaps, at the root of the question, is where does the "extra" mass come from in the cases that are farther apart, where "extra" means the difference between the $2M$ Schwarzschild mass of the merged BH and the sum of the "irreducible masses" pre-merger.  The answer, I think, is that it comes from the EM field energy that is drawn into the final black hole.
