I understand that the quantization of energy (in joules) and charge (in coulombs) are somewhat arbitrary
That's not quantization, they are units of measure.
And whilst they are arbitrary in one sense, they are tied to other measurements of an observable nature. So they form part of a consistent measurement system. Where they have a relationship to constants (for example the speed of light, or absolute temperature, the charge on an electron, etc) then they are tied to physical absolutes.
But it just is confusing to me that we do not have energy inherent in charge.
energy is no more inherent in charge than distance is inherent in time.
energy = voltage * charge
distance = velocity * time
In one case velocity is measured in units of distance per unit time. In the other voltage is measured in energy per unit charge. In both cases they are derived to measure a relationship which is observed in the real world.
To me, both are obvious and intuitive.
If you think of charge in terms of current and time (1C = 1As) then that 1 amp for 1 second can deliver more energy at a higher voltage. That tells you that the coulomb of charge must have more energy. And that energy is described by the voltage.
If you take the time to write down the fundamental units for these things, you'll see they cancel out nicely.
s = ut (s in metres, u in m/s, t in s) we have m = m/s * s
e = VQ (e in kg*m^2/s^2, V in (kg * m^2)/(A * s^3), Q in A*s) we have kg*m^2/s^2 = ((kg*m^2)/(A*s^3)) * (A*s) which also cancels out
For a bit of fun, you can look at the fundamental definition of various units and see where they cancel out. Most often this results in an equation that you are familiar with. However you can also generate equations where you have (say) resistance defined in terms of the universal gravitational constant and other things -- whilst these are true, they may not have any particular use.
If we have a certain amount of "things" that have a certain amount of "elementary charge" surely there is inherent potential energy in that collection of charged things? Why must we additionally introduce energy into the system, in order to obtain voltage?
because, as in the example above, without voltage we cannot relate the two. We have left over units of (in this case) kg, metres, and time. If you look at those leftover units, you'll see mass and distance in the numerator. These tell us that energy is involved (since energy is mass times distance) and the presence of time in the denominator tells us that there is something to do with a rate.
So voltage has something to do with the rate at which we can deliver energy (and that's power).
The rest of the units tell us more about that relationship. (And it turns out that the missing factor is the Amp).
In full, voltage tells us how much power we can deliver at a given current (recall P = V*I).
We know that the coulomb is the unit of charge that is represented by 1 A for 1s, so we can change P = V * I to P = V * (Q/t).
And wasn't Power measured in J/s?. So now we have e/t = VQ/t.
But both of those /t's cancel out, so we are left with e = VQ which is where we started.