In quantum mechanics, the energy of a wave is the number of times it oscillates per second, the momentum of a wave is how many times it oscillates per meter, and the angular momentum of a wave is how many times it oscillates as it goes around a circle. However, these three quantities don't seem to have the right units:
Energy: oscillations per second has units of Hz, not J as is usual for energy
Momentum: oscillations per meter has units of 1/m, not kg m/s as is usual for momentum
Angular momentum: oscillations per circle is dimensionless, but angular momentum has units of kg m^2 / s
##\hbar## is the unit conversion factor that makes everything come out right. It has units of J s = kg m^2 / s
Energy: ##\hbar \times## (something with units of Hz) = something with units of J
Momentum: ##\hbar \times## (something with units of 1/m) = something with units of kg m/s
Angular momentum: ##\hbar \times## (something dimensionless) = something with units of kg m^2 / s
So ##\hbar## is, among other things, the proportionality constant between the usual units of frequency and the usual units of energy. So for any quantum mechanical wave, ##E / \omega = \hbar##. This is just telling us that if we want, we can measure energy in units of Hz. Similarly, we measure momentum in units of 1/m and angular momentum in dimensionless units.
Now let's ask ourselves what are the possible angular momenta in quantum mechanics. As stated above, the angular momentum of a wave in QM is the number of times it oscillates as it winds around a circle. This number has to be an integer, since the wave has to connect back up with itself. So the possible angular momenta are thus 0, 1, 2, 3... . Or including the factor ##\hbar## to make the units more familiar, the possible angular momenta are ##0, \hbar, 2 \hbar, 3\hbar ...##. For technical reasons the half-integer angular momenta 1/2, 3/2, 5/2... are also possible; unfortunately this fact is not straightforward to derive.
We can't know a priori which angular momentum the particles in our universe will have; we just know that it will be an integer or a half-integer (that is, an integer or a half-integer multiple of ##\hbar##). The electron ends up having angular momentum ##1/2 = \hbar/2##, while the photon has angular momentum ##1 = \hbar##.
So we get your equation ##E / \omega = \hbar = 2 \times ## (electron angular momentum).
Or, perhaps a more enlightening way to write this is in units where we measure E in units of Hz, so that ##E = \omega##. Then
##E/\omega = 1 = 2 \times (1/2)##.