I understand the reason for which the wavelengths of the incident and reflected waves must be equal: otherwise, the interference at any fixed position would be constructive at some instants and destructive at others. But why can't two waves of differing amplitude produce a standing wave?
[Physics] Why must traveling waves have the same amplitude to form a standing wave
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Your description of molecular orbital theory is rather misleading, though I concede that it is introduced to students in the way you describe. To really understand what is going on you need a better understanding of how MO theory works.
If we write down the Schrodinger equation for a diatomic molecule like $H_2$ we find it has no analytic solution, so we look for ways of finding approximate solutions. One way of approximating the solution is to write it as a sum of some other functions $\phi_i$:
$$ \Psi = \sum a_i\phi_i $$
We call the functions $\phi_i$ basis functions, and they can be any functions we want - there is no special restriction on what we choose as our basis. However it makes sense to use functions that give a good approximation to $\Psi$ with as few terms as possible. In the case of the $H_2$ molecule an obvious choice for the basis functions is the hydrogen atomic orbitals i.e. the $1s$, $2s$, $2p$, etc.
To get a perfect expression for the $H_2$ wavefunction would require an infinite basis, but we would expect a reasonable approximation with a finite number of terms in the sum. In particular we expect to get a start by considering only two terms i.e. the $1s$ orbitals of the two hydrogen atoms. Let's call these $\phi_1$ and $\phi_2$, so our expression for the molecular wavefunction looks like:
$$ \Psi_{H_2} \approx a_1\phi_1 + a_2\phi_2 $$
And we want to choose the constants $a_1$ and $a_2$ to give the best approximation. For a diatomic molecule this is easy because the molecule is symmetric so $|\Psi|^2$ must be symmetric and that means $|a_1| = |a_2|$. The only possible expressions for $\Psi$ are (give or take a normalising factor):
$$ \Psi_{+} \approx \phi_1 + \phi_2 $$
$$ \Psi_{-} \approx \phi_1 - \phi_2 $$
The energy of these two orbitals is given by:
$$ E = \langle\Psi|H|\Psi\rangle $$
where $H$ is the Hamiltonian for the hydrogen molecule. If you're interested in the details I found quite a nice account here, but assuming the just want an overview the energy depends on the overlap integral $\langle\phi_1|\phi_2\rangle$. In brief, if the overlap integral is large then the energy is low and if the overlap integral is low then the energy is high.
And finally we get to the reason the signs of $\phi_1$ and $\phi_2$ matter. If both are positive then they add up and the sum is large in the region where they overlap. This makes the overlap integral large and gives a low energy. By contrast, if they have different signs then the sum (i.e. the difference) is small in the region where they overlap and the ovelap integral is low and the energy is high.
Take a look at this Desmos animation. Either animate it by clicking the play button next ot the time (t) variable, or drag the slider around to watch the behavior of the waves.
Watch carefully what the standing wave (the black trace) looks like when the waves constructively interfere (I.E. at times when they both look identical) and when the waves destructively interfere (I.E. at times when they looked like mirror images of each other flipped over the y-axis). As you can see, the standing wave is simply the addition of the two traveling waves.
Best Answer
If the travelling waves have the same amplitude then the net rate of transfer of energy at any point is zero and there are stationary positions where the standing wave has zero amplitude - nodes.
In the following animation$^\dagger$ the amplitude of the reflected wave is the same as that of the incident wave.
If the travelling waves are of unequal amplitude then there is a net transfer of energy.
If the amplitudes of the two traveling waves are $A$ and $B$ with $A>B$ then you can think of the superposition of the two travelling waves as being the sum of a standing wave formed by two travelling waves of amplitude $B$ and a travelling wave of amplitude $A-B$.
In this animation the amplitude of the left travelling (incident) wave is larger than that of the right travelling (reflected) wave and so there is a net transfer of energy from left to right.
If you look carefully using a vertical ruler as a marker you will observe positions of maximum displacement and positions of minimum (but not zero) displacement.
The graph bottom left of this video shows this maximum and minimum displacement by overlapping the wave profiles as time progresses.
$^\dagger$ Frames for the animations were made with Mathematica code available here, then combined into GIFs with ImageMagick's
convertcommand.