There is (at least) two kind of validity domain of Euler's $v−e+f=2$ polyhedron formula. One is the "Eulerian" polyhedra, i.e. simply connected polyhedra with simply connected faces (see here). The other is the connected planar graphs. We know, that not every connected planar graph is a skeleton of some polyhedron, only the 3-connected ones, so the second domain seems broader.
On the other hand, according Steinitz's theorem, every 3-connected planar graph is the skeleton of some convex polyhedron. And since every convex polyhedron is Eulerian, the set of polyhedra having 3-connected planar skeleton is a subset of the set of all Eulerian polyhedra. My question is, that is this a proper subset (since the simply connectedness of faces imply the connectedness of the skeleton, this would mean, that there are Eulerian polyhedra with non-planar skeleton ), or every Eulerian polyhedron is combinatorially equivalent to a convex polyhedron? (I call two polyhedra combinatorially equivalent if they have the same skeleton)
So, my real question is:
Can the skeleton of an Eulerian polyhedron be non-planar?
Or, its reverse:
Is it true that every Eulerian polyhedron is combinatorially equivalent to a convex polyhedron?
I think, that "simply connected polyhedra with simply connected faces" is not sufficient for having Euler-Characteristic 2, because there are simply connected polyhedra with simply connected faces that aren't homeomorphic with $S^2$ (see here). It seems, that we should require additionally, that the polyhedra is topologically a manifold (or perhaps equivalently (?), that the polyhedron, and also its interior is connected). I think, that if we define "Eulerian polyhedron" so, then, according the comment of @BalarkaSen, we can state, that every Eulerian polyhedron is combinatorially equivalent to a convex polyhedron.