Why don't non-Hermitian operators with all real-eigenvalues correspond to observables? Suppose you could construct an operator that was non-Hermitian but had all real eigenvalues or could at least be restricted in a way to create only real eigenvalues, why would this operator not correspond to an observable quantity?
 A: Post measurement you want the eigenspaces to be orthogonal (and to have the projection onto the eigenspace to be entangled with a state of the measurement device).
So you want the different eigenspaces to be orthogonal. And you want to be able to evolve to a post measurement state that has the right kinds of states. So really it is about the kinds of end states you can evolve to. So that's what you want, let's talk about why your proposal doesn't get you that.
Just because the eigenvectors are real does not mean that eigenvectors with different eigenvalues are orthogonal.
The orthogonal decomposition is the first essential thing you want. The requirement that the eigenvalues be real isn't as big a deal, you might even want to model decay (poorly) with a complex energy. But you still want the different eigenspaces to be orthogonal.
And you do want more than that. You really want the eigenvectors to contain a maximal orthonormal set, and for a maximal set of commuting operators you get a unique such set, the set of mutual eigenvectors.
And it is after you have this that you can, for instance, finally bring up probability theory (and make a sample space and all that).
Now there is a whole branch of physics (with thousands of papers) based on the operators you mention. But when it comes time to do probability and compute expectation values, they ... change the geometry .. thus making the eigenspaces orthogonal in the new geometry, thus making the operators exactly the normal operators (pun not intended).
So considering non-Hermitian operators gains you nothing in the end. To get the probability part, you pick a geometry where the observables are Hermitian anyway so you could have just worked with that geometry and the Hermitian operators on it. Now, it doesn't necessarily make it wrong if you get the right answers and if doing it in a particular way happens to be computationally easier or otherwise nice then good for you.  But non-orthogonality of the eigenvectors is a problem.
And since I mentioned Hermitian and non-Hermitian operators I do think it is a good time to point out then when you look for a geometry to make the eigenvectors orthogonal you could be doing this because you are aiming for a self-adjoint operator rather than just aiming for a Hermitian operator. Mostly on that front I just wanted you to know that both are out there. And having real eigenvalues won't give you either if the eigenvectors aren't orthogonal.
I haven't seen a paper that says that any self adjoint operator has an experimental realization. So it could be easier to catalog how things don't work rather than to think every nice enough operator can be measured.
A: 1) If all the eigenvalues of an operator are real, then it is Hermitian.  You can see this by writing the operator (call it A) in the eigenvector basis.  Then A has all real eigenvalues along its diagonal and zeros everywhere else.  Therefore, $A^\dagger = A$ which means it is Hermitian.
2) Many of the operators that we call "observables" are the generators of transformations.  For example: J (angular momentum) is the generator of rotations $e^{i\theta J}$, P is the generator of translations $e^{ixP}$, and H is the generator of time translation $e^{itH}$.  If these transformations are to be unitary to conserve probability, then the generator must be Hermitian.  For example unitary means (where $\theta,x,t$ are real):
$$
(e^{ixP})^\dagger e^{ixP}=I
$$
$$
e^{-ixP^\dagger} e^{ixP}=I
$$
$$
-P^\dagger + P =0
$$
therefore P is Hermitian.
