The free theory does not correspond to $\square \phi = 0$ but to the interaction turned off, in the OP's case, the free theory means $g\rightarrow 0$.
Is there something like $|\Omega\rangle$ in QED?
Yes it is as you have written above but where the free theory also contains free fermion and a free abelian gauge boson instead of a scalar sector.
Physically speaking, what would be the difference between $|\Omega\rangle$ and $|0\rangle$?
$\langle \Omega | {\rm e}^{i H T/\hbar} | \Omega\rangle$ is perhaps easier to understand and is usually given diagrammatically by all vacuum diagrams (loops without external legs). In an interacting theory, i.e. where there are interaction terms, the amplitudes need to be normalized against the contributions of all such closed loops (see Figure 1). This you can see from expanding the denominator in Eq. 7.64 of Schwartz's book for example. That is $|\Omega \rangle$ takes into account self-energies and interactions in general, while $|0\rangle$ knows nothing about them.
I once read somewhere (unfortunately I don't remember for reference) that the QED vacuum is the tensor product of the vacuum of free electromagnetism with the vacuum of the free fermionic theory (the unoccupied states of each theory). Is this right?
It is right in spirit but it is somewhat meaningless. From the principles of quantum mechanics, there is the notion that coupling to particles (fields in this case) amounts to simply tensoring the Hilbert spaces of states together. In this sense since QED is built from an electromagnetic sector and a fermion sector, coupling them amounts to describing excitations in some Hilbert space which should formally be some tensor product of the individual ones. If they were not interacting it is somewhat obvious that the tensor product of the ground states is the ground state of the coupled system. However when interactions are turned on, the picture is not so clean to my knowledge.
