The basic idea is the following one. For the sake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time $t$).
On a symplectic $2n$ dimensional manifold (a space of phases), $(M, \omega)$, where $\omega$ is the symplectic 2-form, you first define the Hamiltonian vector field $X_f$ associated to a smooth function $f: M \to \mathbb R$ as the unique vector field such that:
$$\omega(X_f, \cdot) = df\:.\tag{1}$$
The definition is well-posed because $\omega$ is non degenerate by hypotheses. $\omega$ is also antisymmetric and closed by hypotheses, therefore due to a theorem due to Darboux, on a suitable atlas which always exists $\omega = \sum_{i=1}^n dq^i \wedge dp_i$. In this picture you recover the standard $q-p$ formulation of Hamiltonian mechanics.
Next, you have the Poisson bracket of two smooth functions defined as
$$\{f, g\}:= \omega(X_f,X_g)\:.\tag{2}$$
The (generally local) one-parameter group of diffeomorphisms generated by $X_f$
turns out to be made of canonical transformations in the standard sense of Hamiltonian mechanics. $f$ is said to be the Hamiltonian generator of that transformation. Using Darboux' atlas, i.e. coordinates, $q^1,\ldots, q^n, p_1,\ldots, p_n$, both Hamilton equations and Poisson brackets assume the standard form more familiar to physicists.
If $H$ is a preferred function $f$ called the Hamiltonian function, the integral lines of $X_H$ are nothing but the solution of Hamilton equations.
With these definition it turns out that, if $[\:.,\:.]$ is the standard commutator of vector fields,
$$[X_f,X_g] = X_{\{f,g\}}\:.\tag{3}$$
As an immediate consequence of (3), you see that if $\{f,H\}=0$, then the integral lines of $X_H$ remains integral lines of $X_H$ also under the action of the group generated by $X_f$. In this case you have a dynamical symmetry.
Moreover, from (1) and (2), $$X_H(f) = \{f,H\}$$
so that $\{f,H\}=0$ also implies that $f$ is invariant under the Hamiltonian flow, i.e., it is a constant of motion.
The fact that $f$ is a constant of motion and that it generates (canonical) transformations which preserve the evolution of the system are equivalent facts.
This fantastic equivalence does not hold within the Lagrangian formulation of mechanics.
In this scenario, suppose that the $N$ dimensional Lie group $G$ freely acts on $M$ in terms of diffeomorphisms bijective. The one-parameters of the group define corresponding one parameter groups of diffeomorphisms whose generators have the same Lie algebra as that of $G$. So if $e_1,\ldots, e_n$ is a basis of $\mathbb g$ (the Lie algebra of $G$), with
$$[e_i,e_j] = \sum_{k=1}^N c^k_{ij}e_k$$
you correspondingly find, for the associated vector fields defining th corresponding one-parameter groups of diffeomorphisms
$$[X_i,X_j] = \sum_k c^k_{ij}X_k\:.$$
Suppose eventually that each $X_i$ can be written as $X_{f_i}$ for a corresponding smooth function $f_i : M \to \mathbb R$. In this case, the one-parameter group of diffeomorphisms generated by $X_f$ is a one-parameter group of canonical transformations. (This automatically happens when the action of $G$ preserves the symplectic form.)
Consequently,
$$X_{\{f_i,f_j\}} = [X_{f_i},X_{f_j}] = \sum_k c^k_{ij}X_{f_k}\:.$$
Using the fact that the Poisson brakes are bi-linear:
$$X_{\{f_i,f_j\} - \sum_k c^k_{ij}f_k} =0$$
and thus, since $f \mapsto X_f$ is injective up an additive constant to $f$,
$$\{f_i,f_j\} = q_{ij} +\sum_k c^k_{ij}f_k$$
the constants $q_{ij}$ generally appear, sometime (as it happens if $G=SO(3)$) you can re-absorb them in the definition of the $f_i$ which, in turn, are defined up to additive constants. (It is a co-homological problem depending on $G$).
Best Answer
Yes, it is. The phase-space origin in your ill-met infinite-dimensional representations maps to the trivial zero operator in Hilbert space, just like the zero matrix for the 3×3 matrices provided.
Specifically, you have the E(2) brackets, $$ [\pi'(p_1),\pi'(p_2)]=0, \qquad [\pi'(l),\pi'(p_1)]=\pi'(p_2), \qquad [\pi'(l),\pi'(p_2)]=-\pi'(p_1), $$ where your Casimir works; but it is not in the Lie algebra.
In point of fact, your target bracket can also be computed by inspection as a plain commutator of the Schroedinger realization, $$ \pi'(l)= -q_1\partial/\partial q_2 + q_2\partial/\partial q_1 ,\qquad \pi'(p_1)=-\partial/\partial q_1, \qquad \pi'(p_2)=-\partial/\partial q_2, \\ \pi^{'~ 2}(p_1)+ \pi^ {'~ 2}(p_2)= \frac{\partial^2 }{\partial q_1^2}+ \frac{\partial^2 }{\partial q_2^2}~~. $$ Note the last line is not in the Lie algebra, but it is still "represented" faithfully by this "crypto-quantization" realization.