Why exactly does the interaction of a positron and an electron result in energy? Forgive the extremely basic question, but I'd like to know exactly why the collision of these two particles results in the release of energy (EM radiation?)
 A: Remember that mass is already a form of energy ($E=mc^2$) and so electron-positron annihilation doesn't "create" energy, it just transforms it. In general nature prefers to convert mass into other kinds of energy, just as it prefers to convert potential energy into kinetic energy. So for example a free muon will decay into an electron and some other particles:
$$
\mu^{-} \rightarrow e^{-} + \nu_{\mu} + \bar{\nu_e}
$$
Both the left and right sides of the above equation have the same conserved quantum numbers (charge, lepton number, and so on) but the sum of the rest masses on the right hand side is less than the sum on the left, and so the right hand side is preferred.
An isolated electron is stable because in:
$$
e^{-} \rightarrow X
$$
there are no combination of particles you can put in for $X$ that will both have the same quantum numbers as an electron and also lower total rest mass.
Positrons and electrons, however, have opposite quantum numbers, so a system consisting of a positron and an electron ends up having a net charge of 0 (similarly for lepton number and other conserved quantities). So in
$$
e^{-} + e^{+} \rightarrow \gamma + \gamma
$$
the left and right hand side have the same conserved numbers, but the right hand side has less rest mass... so that's the side that nature prefers. Note that you need two photons in order to preserve momentum.
(For the pedantic: there are some details glossed over somewhat in my answer and "nature prefers lower rest mass" is an oversimplification, but the details are probably out of scope for what the OP is looking for.)
A: Perhaps one way to look at this is uncertainty. If the positron and electron existed at the same point in space then there would be no uncertainty in their positions relative to one another By becoming an  electromagnetic radiation wave they are spread out at no particular location.
This would mean black holes cant have infinite density too.
It is also the case that gamma rays crashing into lead become positrons and electrons so reversing this means positrons and electrons become gamma rays
A: The "why" is charge. Charge couples to the electromagnetic field. The reason there is (mostly$^1$) only EM radiation and no matter left is because the positron is the antimatter complement of the electron, so all their conserved quantum numbers are complementary (charge, lepton number), and add to zero.
Also, they don't have color charge, so strong interactions do not occur.
[1] The leptons interact via the weak interaction, so $e^++e^-\rightarrow Z \rightarrow \nu_e + \bar\nu_e $ is possible, but not experimentally detectable, even if the amplitude were large (it is not). See https://arxiv.org/abs/hep-ph/9602382
