I think the misunderstanding at the heart of this question is about what it takes to specify a quantum field theory. A Hilbert space on which a representation of the Poincaré group acts is not enough by itself, you also need a local Hamiltonian with a positive energy spectrum. To see why that is, look at your Hilbert space of (supposedly free) multiparticle states and consider that it wouldn't change at all if you were to introduce a very weak interaction (too weak to create any bound states); but clearly the free theory and the weakly-coupled theory are different (they have different S-matrices for one), so more than the Hilbert space is needed to capture the relevant theory.

(To give a – very vague and potentially misleading, but maybe still helpful – toy analogy: in one-dimensional quantum mechanics, all theories with a purely discrete energy spectrum have Hilbert spaces isomorphic to $\ell^2$, but the harmonic oscillator $V(x)=\frac{\omega^2}{2}x^2$ is a different system from the one with potential $V(x)=\frac{\lambda}{4!}x^4$; the Hamiltonian makes all the difference.)

Where things go wrong if the spin-statistics relation is violated is when you try to construct a local Hamiltonian with positive energy spectrum. You either get a Hamiltonian whose energy spectrum is not bounded from below, or you find that the field operators come out to be identically zero. Neither case corresponds to a quantum field theory.

I think the misunderstanding at the heart of this question is about what it takes to specify a quantum field theory. A Hilbert space on which a representation of the Poincaré group acts is not enough by itself, you also need a local Hamiltonian with a positive energy spectrum. To see why that is, look at your Hilbert space of (supposedly free) multiparticle states and consider that it wouldn't change at all if you were to introduce a very weak interaction (too weak to create any bound states); but clearly the free theory and the weakly-coupled theory are different (they have different S-matrices for one), so more than the Hilbert space is needed to capture the relevant theory.

(To give a – very vague and potentially misleading, but maybe still helpful – toy analogy: in one-dimensional quantum mechanics, all theories with a purely discrete energy spectrum have Hilbert spaces isomorphic to $\ell^2$, but the harmonic oscillator $V(x)=\frac{\omega^2}{2}x^2$ is a different system from the one with potential $V(x)=\frac{\lambda}{4!}x^4$; the Hamiltonian makes all the difference.)

Where things go wrong if the spin-statistics relation is violated is when you try to construct a local Hamiltonian with positive energy spectrum. You either get a Hamiltonian whose energy spectrum is not bounded from below, or you find that the field operators come out to be identically zero. Neither case corresponds to a quantum field theory.

Now that the question has been edited, it seems clear that the issue is something else entirely, namely the concept of locality.

Locality in the context of quantum field theory requires that the Hamiltonian can be written in terms of products of field operators at the same spacetime point only. The single-particle Hamiltonian acting on the single-particle Hilbert space isn't written in terms of field operators at all (the field operators contain the creation and annihilation operators in Fock space, which link different $n$-particle Hilbert spaces, so the field operators don't even act on the single-particle Hilbert space), so the argument "these are 'local' as soon as the 1 particular space 𝐻 we started from is" doesn't work.

I think the misunderstanding at the heart of this question is about what it takes to specify a quantum field theory. A Hilbert space on which a representation of the Poincaré group acts is not enough by itself, you also need a Hamiltonian with a positive energy spectrum. To see why that is, look at your Hilbert space of (supposedly free) multiparticle states and consider that it wouldn't change at all if you were to introduce a very weak interaction (too weak to create any bound states); but clearly the free theory and the weakly-coupled theory are different (they have different S-matrices for one), so more than the Hilbert space is needed to capture the relevant theory.

(To give a – very vague and potentially misleading, but maybe still helpful – toy analogy: in one-dimensional quantum mechanics, all theories with a purely discrete energy spectrum have Hilbert spaces isomorphic to $\ell^2$, but the harmonic oscillator $V(x)=\frac{\omega^2}{2}x^2$ is a different system from the one with potential $V(x)=\frac{\lambda}{4!}x^4$; the Hamiltonian makes all the difference.)

Where things go wrong if the spin-statistics relation is violated is when you try to construct a local Hamiltonian with positive energy spectrum. You either get a Hamiltonian whose energy spectrum is not bounded from below, or you find that the field operators come out to be identically zero. Neither case corresponds to a quantum field theory.

I think the misunderstanding at the heart of this question is about what it takes to specify a quantum field theory. A Hilbert space on which a representation of the Poincaré group acts is not enough by itself, you also need a local Hamiltonian with a positive energy spectrum. To see why that is, look at your Hilbert space of (supposedly free) multiparticle states and consider that it wouldn't change at all if you were to introduce a very weak interaction (too weak to create any bound states); but clearly the free theory and the weakly-coupled theory are different (they have different S-matrices for one), so more than the Hilbert space is needed to capture the relevant theory.

(To give a – very vague and potentially misleading, but maybe still helpful – toy analogy: in one-dimensional quantum mechanics, all theories with a purely discrete energy spectrum have Hilbert spaces isomorphic to $\ell^2$, but the harmonic oscillator $V(x)=\frac{\omega^2}{2}x^2$ is a different system from the one with potential $V(x)=\frac{\lambda}{4!}x^4$; the Hamiltonian makes all the difference.)

Where things go wrong if the spin-statistics relation is violated is when you try to construct a local Hamiltonian with positive energy spectrum. You either get a Hamiltonian whose energy spectrum is not bounded from below, or you find that the field operators come out to be identically zero. Neither case corresponds to a quantum field theory.