This is a good question, and the answer is rather subtle, and I think a physicist and a mathematician would answer it differently.
Mathematically, a Hilbert space is just any complete inner product space (where the word "complete" takes a little bit of work to define rigorously, so I won't bother).
But when a physicist talks about "the Hilbert space of a quantum system", they mean a unique space of abstract ket vectors $\{|\psi\rangle\}$, with no preferred basis. Exactly as you say, you can choose a basis (e.g. the position basis) which uniquely maps every abstract state vector $|\psi\rangle$ to a function $\psi(x)$, colloquially called "the wave function". (Well, the mapping actually isn't unique, but that's a minor subtlety that's irrelevant to your main question.)
The confusing part is that this set of functions $\{\psi(x)\}$ also forms a Hilbert space, in the mathematical sense. (Mumble mumble mumble.) This mathematical Hilbert space is isomorphic to the "physics Hilbert space" $\{|\psi\rangle\}$, but is conceptually distinct. Indeed, there are an infinite number of different mathematical "functional representation" Hilbert spaces - one for each choice of basis - that are each isomorphic to the unique "physics Hilbert space", which is not a space of functions.
When physicists talk about "the Hilbert space of square-integrable wave functions", they mean the Hilbert space of abstract state vectors whose corresponding position-basis wave functions are square integrable. That is:
$$\mathcal{H} = \left \{ |\psi\rangle\ \middle|\ \int dx\ |\langle x | \psi \rangle|^2 < \infty \right \}.$$
This definition may seem to single out the position basis as special, but actually it doesn't: by Plancherel's theorem, you get the exact same Hilbert space if you consider the square-integrable momentum wave functions instead.
So while "the Hilbert space of square integrable wave functions" is a mathematical Hilbert space, you are correct that technically it is not the "physics Hilbert space" of quantum mechanics, as physicists usually conceive of it.
I think that in mathematical physics, in order to make things rigorous it's most convenient to consider functional Hilbert spaces instead of abstract ones. So mathematical physicists consider the position-basis functional Hilbert space as the fundamental object, and define everything else in terms of that. But that's not how most physicists do it.
If we demand that each permissible function must have a decomposition in the Hamiltonian eigenbasis, does this ban some functions that were previously allowed in this Hilbert space?
It does not imply any restriction on the admissible vectors in the Hilbert space.
Every vector in a Hilbert space is a (generally infinite) superposition of eigenvectors of a selfadjoint operator with pure point spectrum. If a Hamiltonian, or every other operator, like a component of the angular momentum, has pure point spectrum this is the case.
For instance, for particles with spin, is every square-integrable wave-function of the form $\psi(x,y,z,s)$ allowed?
Yes, it is allowed. Actually you need $2S+1$ such wavefunctions if the spin is $S$ and the integer $s$ varies from $-S$ to $S$.
Best Answer
There is a certain subtlety to your question.
For quantum systems with a finite number of degrees of freedom, as commonly dealt with in intro QM, things are relatively simple:
Yes and no: the Hamiltonian certainly determines a basis for the Hilbert space of states, but the working Hilbert space depends on the domain of definition of the problem and on associated boundary conditions. See particle in a 3D-box vs. free particle on the entire 3D-space, as well as particle in a box with Dirichlet b.c.-s vs. particle in a box with periodic b.c.-s, etc. Alternatively, the Hilbert space is determined by the algebra of system observables, as pointed out in user1620696's answer, but the two descriptions are eventually equivalent. Moreover, there exists an even deeper equivalence of Hilbert spaces, see point(3) below.
Each particle lives in its own Hilbert space, but the combined interacting system lives in the direct product of individual Hilbert spaces. Again, see relation to algebra of observables as in user1620696's answer.
Leaving aside spin and the spin interactions mentioned by Hosein, generally no, for a finite number of degrees of freedom the Hilbert space does not change under perturbations. According to the Stone-von Neumann theorem, in this case all possible Hilbert spaces are isomorphic to one another (or equivalently, there is a unique irreducible representation of the canonical commutation relations), hence distinguishing one over the other makes no formal difference. At most, the total Hilbert space decomposes into a direct sum of multiple isomorphic copies.
For systems with an infinite number of degrees of freedom, which are the domain of Quantum Field Theory, points (1) and (2) above remain largely valid, but the situation does change drastically in regards to point (3).
The Stone-von Neumann theorem does not hold for quantum fields, and one may find that certain unitary transformations defined on one Hilbert space, constructed around a given Hamiltonian, produce states that are orthogonal to that whole Hilbert space and living in an entirely new, inequivalent state space. This is the case of inequivalent vacua of many QFT Hamiltonians, from condensed matter ones (see boson condensation, superconductivity, etc.) to QCD.
Further, the nature of such inequivalent vacua (or better say, unitarily inequivalent representations of the dynamics) is determined by the nature of interactions between free-fields described by some free-particle Hamiltonian and corresponding state space.
For an idea of what is going on, see for instance Sec. 1.2 of this review on Canonical Transformations in Quantum Field Theory.