Confusion on light absorption by electrons
Depending on how large the value $y$ is, the premise of your question can be wrong.

On the other hand, if you jump between two electronic energy levels where $y$ is on the order of vibrational states, it is entirely possible this extra energy will be stored in vibrations of the molecule on the excited state. This energy is then typically radiated away quite quickly until you reach the ground state of the higher electronic energy level, and then the emission takes place where you could measure the energy $x$ again but in an emission spectrum.

It is entirely possible, however, that the emission process will happen from above the $v=0$ state on the excited surface. It is also possible that you will emit down to different vibrational energy levels on the ground state electronic surface. That is, $0(v=0)\leftarrow1(v=0)$, $0(v=1)\leftarrow1(v=0)$, $0(v=2)\leftarrow1(v=0)$, etc. are all possible and will be observed with varying intensities.

So, in general if you pump in some amount of energy to excite an electron, you don't need to see that same amount of energy coming back out from electronic transitions. You would need to look quite closely at infrared radiation and other energy to make it add up.

As another point, it is actually possible for this energy $x+y$ to be absorbed even if $x+y$ is very far away from the energy of a state. This is how scattering Raman spectroscopy works. You have to think about it differently, however, because the state you are excited to is not a stationary state, but rather a virtual state. See the question I asked here.

So, the short answer is that energy very frequently gets added to other degrees of freedom besides the electronic ones. The same thing goes for vibrational transitions where the excess energy gets put into rotational states. There are rules for how these transitions (and how they couple) will take place.

The other answer, which is less interesting, is that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.