The presumed equivalence between the canonical quantization and the Fock space representation is only a particular case.
The canonical formalism provides only with canonical Poisson brackets. The first step according to Dirac's axioms is to replace the Poisson brackets by commutators and since these commutators satisfy the Jacobi identity, they can be represented by linear operators on a Hilbert space.
Canonical quantization does not specify the Hilbert space.
Finding a Hilbert space where the operators acts linearly and satisfy the commutation relations is a problem in representation theory. This task is referred to as "quantization" in the modern literature.
The problem is that in the case of free fields, this problem does not have a unique solution (up to a unitary transformation in the Hilbert space). This situation is referred to as the existence of inequivalent quantizations or inequivalent representations. The Fock representation is only a special case. Some of the quantizations are called "non-Fock", because the Hilbert space does not have an underlying Fock space structure (i.e., cannot be interpreted as free particles), but there can even be inquivalent Fock representations.
Before, proceeding, let me tell you that inequivalent quantizations may be the areas where "new physics" can emerge because they can correspond to different quantum systems.
Also, let me emphasize, that the situation is completely different in the finite dimensional case. This is because that due to the Stone-von Neumann theorem, any representation of the canonical commutation relations in quantum mechanics is unitarily equivalent to the harmonic oscillator representation. Thus the issue of inquivalent representations of the canonical commutation relations occurs only due to the infinite dimensionality.
For a few examples of inquivalent quantizations of the canonical commutation relations of a scalar field on a Minkowski space-time, please see the following article by: Moschella and Schaeffer. In this article, they construct inequivalent representations by means of Bogoliubov transformation which changes the vacuum and they also present a thermofield representation. In all these representations the canonical operators are represented on a Hilbert space and the canonical commutation relations are satisfied. The Bogoliubov shifted vacuum cases correspond to broken Poincare' symmetries. One can argue that these solutions are unphysical, but the symmetry argument will not be enough in the case of quantization on a general curved nonhomogenous manifold. In this case we will not have a "physical" argument to dismiss some of the inequivalent representations.
The phenomena of inequivalent quantizations can be present even in the case of finite number of degrees of freedom on non-flat phase spaces.
Having said all that, I want nevertheless to provide you a more direct answer to your question (although it will not be unique due to the reasons listed above). As I understand the question, it can be stated that there is an algorithm for passing from the single particle Hilbert space to the Fock space. This algorithm can be summarized by the Fock factorization:
$$ \mathcal{F} = e^{\otimes \mathcal{h}}$$
Where $\mathcal{h}$ is the single particle Hilbert space and $\mathcal{F}$ is the Fock space. As stated before canonical quantization provides us only with the canonical commutaion relations:
$$[a_{\mathbf{k}}, a^{\dagger}_{\mathbf{l}}] = \delta^3(\mathbf{k} - \mathbf{l}) \mathbf{1}$$
At this stage we have only an ($C^{*}$)algebra of operators. The reverse question about the existence of an algorithm starting from the canonical commutation relations and ending with the Fock space (or equivalently, the answer to the question where is the Hilbert space?) is provided by the Gelfand -Naimark-Segal construction (GNS), which provides representations of $C^{*}$ algebras in terms of bounded operators on a Hilbert space.
The GNS construction starts from a state $\omega$ which is a positive linear functional on the algebra $ \mathcal{A}$ (in our case the algebra is the completion of all possible products of any number creation and annihilation operators).
The second step is choosing the whole algebra as an initial linear space $ \mathcal{A}$. In general, there will be null elements satisfying:
$$\omega (A^{\dagger}{A}) = 0$$
The Hilbert space is obtained by identifying elements differing by a null vector:
$$ \mathcal{H} = \mathcal{A} / \mathcal{N} $$
($\mathcal{N} $ is the space of null vectors).
The inner product on this Hilbert space is given by:
$$(A, B) = \omega (A^{\dagger}{B}) $$
It can be proved that the GNS construction is a cyclic representation where the Hilbert space is given by the action of operators on a cyclic "vacuum vector". The GNS construction gives all inequivalent representations of a given $C^{*}$ algebra (by bounded operators). In the case of a free scalar field the choice of a Gaussian state defined by its characteristic function:
$$ \omega_{\mathcal{F} }(e^{\int\frac{d^3k}{E_k} z_{\mathbf{k}}a^{\dagger}_{\mathbf{k}} + \bar{z}_{\mathbf{k}}a^{\mathbf{k}} }) = e^{\int\frac{d^3k}{E_k} \bar{z}_{\mathbf{k}} z_{\mathbf{k}}}$$
Where $z_{\mathbf{k}}$ are indeterminates which can be differentiated by to obtain the result for any product of operators.
The null vectors of this construction will be just combinations vanishing due to the canonical commutation relations (like $a_1 a_2 - a_2 a_1$). Thus this choice has Bose statistics. Also subspaces spanned by a product of a given number of creation operators will be the number subspaces.
The state of this specific construction is denoted by:
$\omega_{\mathcal{F}}$, since it produces the usual Fock space. Different state choices may result inequivalent quantizations.
It's just a convention to write the Dirac Hamiltonian in terms of electron and positron operators. The operator $b_p^{s\dagger}$ creates a "hole", i.e. a positron, which is the same as annihilating an electron. So we could just as well define $c_p^s = b_p^{s\dagger}$ and write the Hamiltonian in terms of $a$ and $c$ instead of $a$ and $b$. Then everything is expressed in terms of electron operators.
So we are reduced to asking why you need two sets of operators $a$ and $c$. The reason is simply that the dispersion relation of the Dirac equation is $E = \pm \sqrt{p^2 + m^2}$, so at any given momentum there are two bands. Hence, you need one operator that creates/annihilates an electron in the top band, and one which creates/annihilates an electron in the bottom band. This would equally well be true in a condensed matter system treated in momentum space, if there are multiple bands.
(As for why this convention is chosen: in relativistic quantum theory, if you write everything in terms of electrons, you are forced to conclude that the vacuum $|0\rangle$ has all the negative-energy states occupied, since $c_p^{s\dagger}|0\rangle = 0$. This was, in fact, Dirac's original picture, but conceptually it's easier to imagine that the vacuum contains no particles. On the other hand, in condensed matter physics, we are used to thinking about our solid-state materials as containing a lot of electrons, so it's actually less confusing to take the opposite point of view.)
Best Answer
''They [...] showed that Bogoliubov transformation with infinity space volume yields unitary inequivalent representations.'' Yes, this is an important effect and the source for phenomena like phase transitions and superconductivity. The isotropic limit of infinite space volume is called the thermodynamic limit. It figures everywhere in the derivation of classical thermodynamics from classical or quantum statistical mechanics and erases all surface effects.
Without thermodynamic limit, there would be no phase transitions! Statistical mechanics of a system with finitely many particles always leads to an equation of state without discontinuities in the response functions. The latter (i.e., the phase transitions in the sense of thermodynamics) appear only in the thermodynamic limit. (Indeed, a system can be defined as being macroscopic if the thermodynamic limit is an appropriate idealization. Note that the Avogadro number $N$ is well approximated by infinity.)
The equivalence with ordinary quantum mechanics suggested by the derivation of second quantization only holds at a fixed number of particles. But quantum field theory is the formulation at an indefinite number of particles. This already requires (even without the thermodynamic limit and independent of surface effects) a Hilbert space with an infinite number of degrees of freedom, where the canonical commutation relations have infinitely many inequivalent unitary representations.