Let me first answer your question "is it wrong to consider topological superconductors (such as certain p-wave superconductors) as SPT states? Aren't they actually SET states?"
(1) Topological superconductors, by definition, are free fermion states that have time-reversal symmetry but no U(1) symmetry (just like topological insulator always have time-reversal and U(1) symmetries by definition). Topological superconductor are not p+ip superconductors in 2+1D. But it can be p-wave superconductors in 1+1D.
(2) 1+1D topological superconductor is a SET state with a Majorana-zero-mode at the chain end. But time reversal symmetry is not important. Even if we break the time reversal symmetry, the Majorana-zero-mode still appear at chain end. In higher dimensions, topological superconductors have no topological order. So they cannot be SET states.
(3) In higher dimensions, topological superconductors are SPT states.
The terminology is very confusing in literature:
(1) Topological insulator has trivial topological order, while topological superconductors have topological order in 1+1D and no topological order in higher dimensions.
(2) 3+1D s-wave superconductors (or text-book s-wave superconductors which do not have dynamical U(1) gauge field) have no topological order, while 3+1D real-life s-wave superconductors with dynamical U(1) gauge field have a Z2 topological order. So 3+1D real-life topological superconductors (with dynamical U(1) gauge field and time reversal symmetry) are SET states.
(3) p+ip BCS superconductor in 2+1D (without dynamical U(1) gauge field) has a non-trivial topological order (ie LRE) as defined by local unitary (LU) transformations. Even nu=1 IQH state has a non-trivial topological order (LRE) as defined by LU transformations. Majorana chain is also LRE (ie topologically ordered). Kitaev does not use LU transformation to define LRE, which leads to different definition of LRE.
Best Answer
A symmetry-protected topological phase has a certain symmetry. Any Hamiltonian in this phase can be adiabatically deformed (i.e. without closing the gap) into a Hamiltonian whose ground state is a product state, but the symmetry must be explicitly broken during the deformation process and then restored at the end. As a visually analogy, there is a "wall" crossing the submanifold of parameter space that respects the symmetry, and the wall separates the SPT phase from the totally trivial phase with a product ground state. But if you are allowed to temporarily break the symmetry, then you can leave the submanifold and "jump over the wall" before ending up back in the submanifold and restoring the symmetry.
A Hamiltonian in a topologically ordered phase cannot be deformed into a Hamiltonian with a product ground state by any means whatsoever (without closing the gap). Here, the "wall" crosses the entire parameter space of all possible (local) perturbations (it's infinitely high and can't be jumped over). The phase does not need to have any symmetry. This is much stronger condition.
The two concepts are closely related mathematically as well. It turns out that topologically ordered states are much more exotic than SPT states. (E.g. they have "anyonic" excitations with neither bosonic nor fermionic exchange statistics, while SPT's do not. At least, not in the bulk - things get a little subtle at the boundary.) But if you mathematically "gauge" the symmetry that protects the SPT, then you get a theory that is morally very similar to a topologically ordered state. Also, both types of systems can usefully by classified using cohomology theory.