It is not a postulate that the Hilbert space for QFT is a Fock space. In fact, for interacting theories is often almost surely not a Fock space. The requirements for a Hilbert space to be the space of a QFT is that the Wightman axioms are satisfied.
For free theories, a suitable Fock representation of the canonical commutation relations satisfies the Wightman axioms (in the sense that there exist a vacuum state, the unitary representation of the Poincaré group etc. in the corresponding Fock space); and this is why the Fock space is commonly introduced - also its definition is in my opinion quite intuitive. The other representations used in interacting theories, when known, are rather abstract, and it may even not be clear if they are set in a Fock space or not (almost surely not, for interesting theories).
Let me expand the ideas a bit more, related to the last part of your question. When we write the canonical commutation relations for quantum fields, it is customary to start from a suitable "one-particle" space $\mathfrak{h}$ (at least it was called like that by Irving Segal). Such space is e.g. the Hilbert space on which you build the Fock space $\Gamma_{s/a}(\mathfrak{h})$ (as direct sum of tensor products), and it encodes roughly speaking the properties of the related classical field (scalar/vector/spinor field,...). However, contrarily to what happens in quantum mechanics, there are infinitely many unitarily inequivalent irreducible representations of the canonical commutation relations $\mathrm{C}^*$-algebra built on such one-particle space $\mathfrak{h}$. One of those representations is the Fock one, the others still encode the commutation relations of the creation/annihilation operators $\{a^{\#}(h),h\in\mathfrak{h}\}$, but such operators may not behave as in the Fock representation: in particular, the intersection of their domains of definition $\bigcap_{h\in\mathfrak{h}}D(a(h))$ may fail to be a dense subset of the Hilbert space.
The following is summarized from two notes (both can be found here) by Prof. Michael Stone (@mikestone). Confusing typos in these notes are corrected.
From a general perspective, the two set of operators $\{a\}, \{b\}$ are two realizations of the CCR/CAR. By the Stone-von Neumann theorem, there is a unitary operator $\mathcal{U}$ that acts on the many-body Hilbert space such that
$$
b_i = \mathcal{U} a_i \mathcal{U}^\dagger
\ \Rightarrow \
b^\dagger_i = \mathcal{U} a^\dagger_i \mathcal{U}^\dagger
$$
Then we find $|0_b\rangle \propto \mathcal{U}|0_a\rangle$, since
$$
a_i |0_a\rangle
= \mathcal{U}^\dagger b_j \mathcal{U} |0_a\rangle = 0
$$
However, as noted by Prof. Stone, the general form of $\mathcal{U}$ is very complicated. Thus we take another approach.
Assume that $U$ is invertible (this is not an "innocent assumption"). Then $b_i |0_b \rangle = 0$ is equivalent to
$$
0 = U^{-1}_{ij} b_j |0_b\rangle
= (a_i - S_{ij} a^\dagger_j) |0_b \rangle,
\quad
S \equiv - U^{-1} V
$$
Using the constraint $U V^\mathsf{T} = \eta V U^\mathsf{T}$, one can show that
$$
S^\mathsf{T} = \eta S
$$
i.e. the matrix $S$ is symmetric for bosons, and anti-symmetric for fermions. Then we construct the bilinear
$$
Q \equiv \frac{1}{2} \sum_{i,j}
S_{ij} a^\dagger_i a^\dagger_j
$$
and use the BCH formula to calculate
$$
\begin{equation*}
e^Q a_i e^{-Q}
= a_i + [Q,a_i]
+ \frac{1}{2}[Q,[Q,a_i]]
+ \cdots
\end{equation*}
$$
Fortunately, the commutator
$$
[Q, a_i] = - S_{ij} a^\dagger_j
$$
commutes with $Q$ (for both bosons and fermions). Therefore we simply get
$$
\begin{equation*}
e^Q a_i e^{-Q}
= a_i + [Q,a_i]
= a_i - S_{ij} a^\dagger_j
\end{equation*}
$$
and $b_i |0_b \rangle = 0$ is equivalent to
$$
e^Q a_i e^{-Q} |0_b\rangle = 0
$$
This can be satisfied by all $a_i$ if and only if
$$
e^{-Q} |0_b \rangle \propto |0_a \rangle
$$
The "if" is obvious. To show "only if", suppose that $|\psi\rangle \equiv e^{-Q} |0_b \rangle \ne |0_a \rangle$; then $e^Q$, which consists of only $\{a^\dagger_i\}$, cannot annihilate $|\psi\rangle$. Finally, up to a normalization constant $\mathcal{N}$, the vacuum $|0_b\rangle$ is
$$
|0_b\rangle = \mathcal{N} e^Q |0_a \rangle
$$
(When I have time I will update on finding $\mathcal{N}$.)
Best Answer
The answer is yes for finitely many operators - this is the content of the Stone-von Neumann theorem: Up to unitary equivalence, there is only a single unitary representation of the CCR $[x_i,p_j] = \delta_{ij}$ (or rather their more well-behaved cousins, the Weyl relations), and it is given by $L^2(\mathbb{R}^n)$ with $n$ the number of position (or momentum) operators and position as multiplication and momentum as differentiation (or vice versa via Fourier transform).
The answer is no for infinitely many operators - this is the content of Haag's theorem: There are many different unitary representations of an infinite collection of CCRs, which is a notoriously annoying issue for rigorous attempts at quantum field theory.