Let me try to answer this question. I apologise if my notation is slightly different, since I will work in some more generality, since the equivariance properties of the creation and annihilation operators are actually more transparent, I believe, relative to the the general linear group instead of the orthogonal group. Also the fact that this is the harmonic oscillator is a red herring. In the case of the harmonic oscillator, we have to introduce more structure, reducing the group of symmetries.
It is also in this case, where the grading to be discussed below coincides (up to a choice of scale) with the energy of the system.
Let $E$ be an $n$-dimensional real vector space and let $E^*$ denote its dual. Then on $H = E \oplus E^* \oplus \mathbb{R}K$ one defines a Lie algebra by the following relations:
$$ [x,y]= 0 = [\alpha,\beta] \qquad [x,\alpha] = \alpha(x) K = - [\alpha,x] \qquad [K,*]=0$$
for all $x,y \in E$ and $\alpha,\beta \in E^*$. This is called the Heisenberg Lie algebra of $E$, denoted $\mathfrak{h}$.
The automorphism group of $\mathfrak{h}$ is the group $\operatorname{Sp}(E\oplus E^*)$ of linear transformations of $E\oplus E^*$ which preserve the symplectic inner product defined by the dual pairing:
$$\omega\left( (x,\alpha), (y,\beta) \right) = -\alpha(y) + \beta(x).$$
Let $\mathfrak{a} < \mathfrak{h}$ denote the abelian subalgebra with underlying vector space $E \oplus \mathbb{R}K$. One can induce a $\mathfrak{h}$-module from an irreducible (one-dimensional) $\mathfrak{a}$-module as follows. Let $W_k$ denote the one-dimensional vector space on which $E$ acts trivially and $K$ acts by multiplication with a constant $k$. Then letting $U$ be the universal enveloping algebra functor, we have that
$$ V_k = U\mathfrak{h} \otimes_{U\mathfrak{a}} W_k$$
is an $\mathfrak{h}$-module. The Poincaré-Birkhoff-Witt theorem implies that $V_k$ is isomorphic as a vector space to the symmetric algebra of $E^*$, which we may (as we are over $\mathbb{R}$) identify with polynomial functions on $E$.
The subgroup of $\operatorname{Sp}(E\oplus E^*)$ which acts on $V_k$ is the general linear group $\operatorname{GL}(E)$ and hence $V_k$ becomes a $\operatorname{GL}(E)$-module. In fact, $V_k$ is graded (by the grading in the symmetric algebra of $E^*$ or equivalently the degree of the polynomial):
$$V_k = \bigoplus_{p\geq 0} V_k^{(p)}$$
and each $V_k^{(p)}$ is a finite-dimensional $\operatorname{GL}(E)$-module isomorphic to $\operatorname{Sym}^p E^*$.
Every vector $x \in E$ defines an annhilation operator: $A(x): V_k^{(p)} \to V_k^{(p-1)}$ via the contraction map
$$E \otimes \operatorname{Sym}^p E^* \to \operatorname{Sym}^{p-1} E^*$$
whereas every $\alpha \in E^*$ defines a creation operator: $C(\alpha): V_k^{(p)} \to V_k^{(p+1)}$ by the natural symmetrization map
$$E^* \otimes \operatorname{Sym}^p E^* \to \operatorname{Sym}^{p+1} E^*.$$
Both of these maps are $\operatorname{GL}(E)$-equivariant, and this is perhaps the most invariant statement I can think of concerning the creation and annihilation operators.
I think the best way to answer this question is to direct you to an introductory textbook on the mathematics of quantum mechanics. One book I really like is Hilbert Space Operators in Quantum Physics by Blank, Exner, and Havlicek.
It sounds like the issue you are dealing with is that the canonical commutation relations can't be satisfied by bounded operators. You need to go to unbounded self-adjoint operators, which means they are only defined on a dense subspace of the Hilbert space. That's just the nature of the beast. In the C*-algebra approach we focus attention on C*-algebras of bounded operators to which the unbounded observables are affiliated.
Maybe part of your question is that if $P$ and $Q$ are only densely defined then $e^{itQ}$ and $e^{isP}$ can only be densely defined too? No, when you exponentiate an unbounded self-adjoint operator you get a bounded, and everywhere-defined, unitary operator. An easy way to do this is to use spectral theory to turn your self-adjoint operator into a multiplication operator, say multiplication by a possibly unbounded real-valued function $f$ on $L^2(X,\mu)$, and then define the exponential of this multiplication operator to be multiplication by the bounded function $e^{itf}$. Again, you need to read a good introductory textbook to learn about this.
Best Answer
I am not sure it is relevant exactly, or if you still have interest in this problem, but perhaps you might find this comment useful or interesting.
$\mathrm{Hom}(\mathbb{Z}^r,G)$ is generally NOT path-connected. For instance, if $r\geq 3$ and $G=\mathsf{SO}(n)$, over $\mathbb{R}$ or $\mathbb{C}$, for $n\geq 4$ then it is disconnected. Thus, in this setting, there are commuting collections $\lbrace A_i\rbrace$ and $\lbrace B_j\rbrace$ that are NOT connected by a path through commuting elements.