Superposition of polarized photon I understand that a $+45^\circ$ polarized photon is a superposition of $0^\circ$ (vertical) and $90^\circ$ (horizontal): $|\!+\!45\rangle = 0.707 (|H\rangle + |V\rangle)$. So measuring the polarization of the photon (using a polarization beam splitter, for example) in the $HV$ basis will give a random result - either $|H\rangle$ or $|V\rangle$. On the other hand, measuring the same polarization in the diagonal ($\pm45^\circ$) basis will give a definitive result: $|\!+\!45\rangle$. It looks to be a equal or balanced superposition only of HV basis.
Now, for an entangled photon, the result of a measurement on the first photon will always be random, regardless of the orientation of the PBS. That is, if measured in the $HV$ basis, the result will be randomly $H$ or $V$. If measured in the diagonal basis, the result will randomly be $|\!+\!45\rangle$ or $|\!-\!45\rangle$. This is true of all angles of the PBS. It looks to have an equal or balanced superposition in all or any basis. 
So, my question is this: Is there a difference in the 'superposition' of the entangled photon versus the polarized photon?
 A: There is a difference, and it is the difference between a quantum superposition and a classical mixed state.
In the case of the single 'superposition' photon, you have what is called a pure state.  As you said, this gives you a 50/50 probability when measured in the $|\pm45\rangle$ basis, but it is a Schrodinger's Cat type of state where the reality is that it actually is in both states at the same time.  This can be proven through interference experiments (for example a Mach-Zender interferometer).
In the case of the 'entangled' photon, we can never talk about the photon by itself, we must consider the whole system of what it is entangled with.  If it is entangled with another photon (perhaps they were created together via parametric down-conversion), then you might for example have a system of two photons that always have different polarizations as $\frac{1}{\sqrt{2}}(|HV\rangle+|VH\rangle)$ (where this denotes a 50% probability of finding the first photon in H and the second photon in V, and a 50% probability of finding the first photon in V and the second photon in H), or you might have two photons that always have the same polarization as $|HH\rangle+|VV\rangle$ (where this denotes a 50% probability of finding both photons as H, and a 50% probability of finding both photons as V).  You can also have any other amplitude ratio or phase relationship between the two.  The point is just that the entanglement tells you there is some correlation between the two photons, without telling you what the state of either one is individually.
So what happens if you measure just one of the entangled photons?  Let's assume they start in the entangled state $|HH\rangle>+|VV\rangle$ and then you measure just the first photon.  Well you still have a 50/50 chance of measuring either $|H\rangle$ or $|V\rangle$, but you most definitely cannot assert that the single photon was in the superposition $|H\rangle+|V\rangle$. You can prove this because, as you said, it also has a 50/50 probability in the $|\pm 45\rangle$ basis, or any other basis, and this is not the way the $|H\rangle+|V\rangle$ superposition behaved.  What you have behaves exactly like a completely unpolarized photon, where its polarization is just random in the classical sense.  And in fact what you have is a completely unpolarized photon, as long as you are considering just the single particle, and not the full two-photon system.
So, the two photon system was in a quantum mechanical pure state, but if you lose one of those photons, the remaining single photon subsystem is a classical mixed state.
In fact, for the example state $|HH\rangle+|VV\rangle$ the only information about the polarization was in the two-photon correlations.  This is called a maximally entangled state.  It is possible to have ensembles (repeated experiments prepared in the same way) that that are partially polarized as well, and the best way to write these types of states is not with the Dirac bra-ket notation that is only good at expressing pure states, but rather with a matrix called a density matrix.
