The multiplication by a phase of the wave function commutes with the
action of the Galilean group.
It is always possible to add a generator, commuting with a Lie algebra generators, to form a Larger Lie algebra. In this case, the larger Lie algebra is called a central extension of the former. The origin of the name is that the added generator (or generators) commute with all the Lie algebra generators, thus they belong to the Lie algebra's center.
For example, The Heisenberg-Weyl algebra:
$$[x, p] = i \hbar \mathbb{1}$$
is a central extension of the two dimensional translation algebra
$\mathbb{R}^2$:
$$[x, p] =0$$
If the central element does not appear in any commutator of the original
Lie algebra generators, the central extension is trivial (all the
$b_{ij}$s are zero) and the algebra is just a direct sum of the two algebras. This is the case in the specific example of the extension described in Ballentine's book. However, this is not the case in the Heisenberg-Weyl algebra, where the commutator of $x$ and $p$ produces the central element. In this case, the central extension is not trivial.
However, this is not the whole story yet. Ballentine is preparing the
backgound for the description of a nontrivial central extension of the Galilean group:
It turns out that the Galilean algebra does not close in neither classical or quantum mechanics without a nontrivial central extension. The Poisson brackets in classical mechanics and the commutator in quantum mechanics of boosts and momenta is not trivial and has the form:
$$[G_i, P_j] = m \delta_{ij}$$
where: $m$ is the particle's mass. Note that the corresponding commutator in the Galilean algebra is vanishing. This result is due to V. Bargmann.
In the classical case, it can be seen quite easily. The Poisson brackets of the Noether charges computed from the free particle Lagrangian corresponding to the boosts and the momenta just satisfy the above relation and they do not Poisson-commute as in the unextended Galilean group algebra.
Finally, let me note that the central element is always represented by a unit matrix in an irreducible representation and a representation of the centrally extended algebra is called a ray representation of the original algebra.
Your picture isn't quite right. In QFT what was the wave function gets promoted to observable operator, and $\mathbf{x}$ gets demoted to parameter on the same level as $t$. The time evolution of an operator is not given by $\operatorname{e}^{-iHt}$, that's the evolution of a state vector. Operators evolve, in the Heisenberg picture, according to: $$\psi(t,\mathbf{x}) = \operatorname{e}^{-iHt} \psi(0,\mathbf{x}) \operatorname{e}^{iHt}.$$ You can go further than that, though, and add in the space translation generators to get:
$$\psi(t,\mathbf{x}) = \operatorname{e}^{-i P_\mu x^\mu} \psi(0) \operatorname{e}^{i P_\mu x^\mu},$$ in the $(+,-,-,-)$ signature metric.
What's going on here is that most treatments of QFT elid over the state vector necessary for a Schrodinger treatment. That state vector still obeys a Schrodinger type equation, it just has to be cast in terms of functional analysis instead of ordinary calculus.
As an example, the free real scalar field has Lagrangian density: $$ \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{m^2}{2} \phi^2.$$ The momentum canonically conjugate to $\phi$ is $\pi \equiv \frac{\partial\mathcal{L}}{\partial \dot{\phi}} = \dot{\phi}$. This produces a Hamiltonian in the usual way: $$H = \int \operatorname{d}^3 x \left[\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{m^2}{2} \phi^2\right].$$ The fields are now promoted to operators that obey the equal time canonical commutation relations, $[\phi(\mathbf{x}), \pi(\mathbf{y})] = i \delta(\mathbf{x}-\mathbf{y})$. The state vector now has to assign a probability density, per unit function space volume ($[\mathcal{D} \phi]$), to every distinct configuration the field can take at any given time. This is known as the wave functional, denoted $\Psi[\phi]$. The canonical commutation relations imply that $\Psi$ obeys a Schrodinger type equation:
$$\int \operatorname{d}^3 x \left[-\frac{1}{2} \frac{\delta^2 \Psi}{\delta \phi(\mathbf{x})^2} + \frac{1}{2}(\nabla\phi)^2 \Psi + \frac{m^2}{2} \phi^2 \Psi \right] = i \frac{\partial \Psi}{\partial t}.$$ This equation is, of course, just the simple harmonic oscillator for which we can construct raising and lower operators in the usual way (after changing to Fourier space). The ground state is given by the Gaussian functional: $$\Psi_0[\phi] \propto \exp\left(-\frac{1}{2}\int \operatorname{d}^3k \left[[\phi(k)]^2 \sqrt{\mathbf{k}^2 + m^2}\right]\right), $$ with excited states built using raising operators, $a^\dagger(\mathbf{k}) = \sqrt[4]{\frac{k^2 + m^2}{4}}\left[\phi(\mathbf{k}) - \frac{i}{\sqrt{k^2 + m^2}} \pi(\mathbf{k})\right]$, in the usual way.
I can only speculate that QFT isn't taught this way in most textbooks for two reasons. First, QFT is primarily used for calculating scattering amplitudes, and other formalisms are easier to get results from. Second, the infinities that plague QFT, requiring renormalization, could be even more difficult to manage in this formalism. This 1996 paper by Long and Shore is one example of professionals using this formalism.
Best Answer
First, note that on the Hilbert space of physical states is realized the projective representations of the Poincare group, precisely, the representations "up to the sign". This can be incorporated automatically by enlarging the Poincare group up to its universal covering group, which is $ISL(2,C)$.
Here is the general statement which is the point: one-particle representations of the $ISL(2,C)$ with non-zero mass $m$ and a spin $s$ are realized by the spinor tensors $ \psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}, \ \frac{m+n}{2}=s$ satisfying the equations $$ \tag 1 \begin{cases}(\partial^{2}+m^{2})\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = 0, \\ \partial^{a\dot{b}}\psi_{aa_{1}...a_{n-1}\dot{b}\dot{b}_{1}...\dot{b}_{m-1}}= 0, \quad \partial^{a\dot{b}} \equiv \partial^{\mu}\sigma_{\mu}^{a\dot{b}}, \ \ \text{for } A,B> 1 \end{cases} $$ As the reminder, the undotted indices correspond to the $SL(2,C)$ group (which is the universal covering group for the Lorentz group $SO(3,1)_{\uparrow}$) transformation $N$, while the dotted indices correspond to the complex conjugated transformation $N^{*}$. Finally, $\sigma_{\mu} \equiv (1, \sigma)$, with $\sigma$ being the three Pauli matrices.
You should understand $(1)$ as the equations defining $\psi$ as the function corresponding to the representation on which two Poincare's group Casimir's operators $\hat{P}^{2}$ (translation/4-momentum squared) and $\hat{W}^{2}$ (Pauli-Lubanski squared) have the values $m^{2}$ and $-m^{2}s(s+1)$ correspondingly. This can be shown by expressing these Casimir's invariants in terms of the spinor tensors objects.
Let's denote the tensor $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ by the representation $\left( \frac{n}{2},\frac{m}{2}\right)$ of the $SL(2,C)$ group to which it corresponds. For the spin one half, from the previous statements we have that both of the representations $\left( \frac{1}{2},0\right)$ and $\left( 0, \frac{1}{2}\right)$ are valid; they're given by the two-component spinors. But corresponding equations, which are simply the Klein-Gordon equations, are far from the Dirac equation... So what's the another point?
In fact, neither $\left( \frac{1}{2}, 0\right)$ nor $\left(0,\frac{1}{2} \right)$ describes the theory which is invariant under $P-,T-,C-$ discrete symmetries separately; from this in particular follows that it's impossible to construct the EM interaction theory of these representations with the static interaction law coinciding with the Coulomb law.
In order to get invariant theory, we need to take the direct sum $$ \tag 2 \left( \frac{1}{2},0\right) \oplus \left( 0,\frac{1}{2}\right) $$ of these representations.
But this direct sum is of course not irreducible representation. In order to make it irreducible (and simultaneously $P-,T-,C-$ invariant), we need to construct a poincare- and $P-,T-,C-$covariant operator which allocates the irreducible representation when acting on the direct sum $(2)$. It's not hard to construct such operator (assuming you can derive $(1)$), and it turns out that it coincides with the Dirac equation operator.
Similar story is true when we consider photons and gravitons. In fact, the massless representations of the Poincare group are characterized by the helicity value, which is the operator's $\hat{h} = \frac{\hat{\mathbf{W}}\cdot \hat{\mathbf P}}{|\hat{\mathbf P}|}$ value, and physically is the projection of the total angular momentum on the direction of motion. Again, the representations with separate values of helicity (say, 1, or -1) are not $P-,T-,C-$invariant. In order to have invariant theory, it's necessary again to take the direct sum, but now without any projectors. In the case of the photons we'll arrive to Maxwell equations on the strength tensor $F_{\mu\nu}$, while in the case of the gravitons we'll arrive to equations on the linearized source-free general relativity Weyl tensor.