Derivation of $E=h\nu$?

I found this question here and the answers got me thinking.

Take a case where $\Psi(x,t)$ is the linear combination of two eigenvectors of a charged particle:

$\Psi(x,t)=c_1\psi_1e^{-iE_1t/\hbar}+c_2\psi_2e^{-iE_2t/\hbar}$.

At $t=0$ the wave function is:

$\Psi(x,0)=c_1\psi_1+c_2\psi_2$.

The probability density distribution is:

$P(x,t)=\Psi^*(x,t)\Psi(x,t)$.

$P(x,t)=|c_1\psi_1|^2+|c_2\psi_2|^2+c_1c_2^*\psi_1\psi_2^*e^{-i(E_1-E_2)t/\hbar}+c_1^*c_2\psi_1^*\psi_2e^{-i(E_1-E_2)t/\hbar}$.

So $P(x,t)$ is (obviously) not constant but contains elements that oscillate with angular frequency $\omega=(E_1-E_2)/\hbar$ and frequency $\nu=(E_1-E_2)/h$.

Does this explain Bohr's equation: $E_1-E_2=h\nu$?

As soon as you write the first equation, you're asserting that $h$ or $\hbar$ is the conversion between frequency and energy, otherwise you wouldn't be able to write $E t / \hbar$ and have it make sense, dimensionally. Therefore, when you take an energy difference, you still use $h$ to get $\nu$ from $E$. What you've called "Bohr's formula" is just the statement that the difference between two atomic energy levels is $h \nu$ for some $\nu$. I suspect the actual information you want to prove is that this is then the frequency of an absorbed or emitted photon, but that just follows from conservation of energy. The real leap is just knowing that there exists a constant, called $h$ or $\hbar$, that can convert between frequency and energy.
As pointed out in zeldredge's answer, what is missing is the connection to the photon. However, I do think you can fix this problem. If you couple a system that exhibits natural oscillations at certain frequencies to an external force that oscillates at some frequency $\omega$, then you get a resonance when $\omega$ matches one of the natural frequencies. Conservation of energy then yields that in a quantum mechanical treatment of the force field, the quanta that appear must have energies of the form $\hbar\omega$, which is indeed the case for a harmonic oscillator.