In the process of going through more posts to ask this question, I think I found a mostly satisfactory way to answer my question.
I'm therefore posting this as an answer because if I put it all into the question, the post would end up being a "is this reasoning correct" kind of post, which I don't like very much.
At the same time, while confident, I'm still not entirely certain that this is the correct way to think about it, which is why I'm posting the whole thing (on top of it being possibly useful for future reference).
This answer to Normal Order of Normal Order, as well as this answer to How exactly is "normal-ordering an operator" defined? go into a somewhat more rigorous explanation of what "applying the CCR" might mean.
My reading of these answers is that most of the apparent oddities in the expressions involving orderings and creation/annihilation operators stem from there really being two wholly different kinds of objects involved:
On the one hand, we can understand the expressions as statements about elements of some free algebra generated by $a$ and $a^\dagger$ (in other words, considering the symbols "$a$" and "$a^\dagger$" as two generic elements of an algebra that do not have any special relation with each other).
For example, $N(aa^\dagger)=a^\dagger a$ is one such case. Here $a$ and $a^\dagger$ don't need to be interpreted as "operators" at all. They're just two symbols, and $N$ is an operator taking expressions and moving all the $a$ on the right.
For the same reason, at this level, $aa^\dagger \neq a^\dagger a+ 1$, because $aa^\dagger$ and $a^\dagger a+1$ are simply two different elements of the free algebra generated by $a,a^\dagger$.
Analogously, considering again the example of this post, it is clear that $N([a,a^\dagger])=0$ holds, because $aa^\dagger-a^\dagger a\neq 1$, and $N(aa^\dagger)=N(a^\dagger a)=a^\dagger a$.
Things change drastically when the CCRs are enforced. This effectively means to take the expressions modulo the equivalence relation generated by $aa^\dagger-a^\dagger a=1$.
Let's denote with $\Pi$ the map sending algebra elements (i.e. generic expressions with $a$ and $a^\dagger$) to the corresponding equivalence classes. Then we have by definition $\Pi(aa^\dagger-a^\dagger a)=\Pi(1)$, $\Pi(aa^\dagger)=\Pi(a^\dagger a + 1)$, etc.
In this language, we can't have a normal ordering operator acting after $\Pi$ has been applied. It simply doesn't make any sense to do so: $\Pi$ sends expressions into equivalence classes of expressions, and $N$ can't be defined on those. It's pretty easy to see why that is so: assuming only linearity you'd get for example
$$N(\Pi( a a^\dagger) ) = N(\Pi(a a^\dagger(a a^\dagger - a^\dagger a)))
= N(\Pi(a a^\dagger a a^\dagger)) - N(\Pi(a a^\dagger a^\dagger a))
\\
= N(\Pi(a^{\dagger 2} a^2)) - N(\Pi(a^{\dagger 2} a^2))
=0 .$$
My point here being that we can't meaningfully apply $N$ after having identified expressions according to the CCRs.
Finally, expressions involving the action of creation and annihilation operators on Fock states (and thus any other ket) necessarily operate at the second level above. That is because those expressions implicitly assume that $a$ and $a^\dagger$ are operators acting on basis states in a specific way. And that automatically imposes $[a,a^\dagger]=1$.
As a concrete example, let me consider the expression $N(e^{-a^\dagger a})=|0\rangle\langle0|$ proved here.
Part of the rationale behind me asking this question stemmed from me trying to answer the question: "does $N(e^{-a^\dagger a})=|0\rangle\langle0|$ mean that $|0\rangle\langle0|$ is normally ordered?".
And in the light of the above discussion, I think I can answer this by saying that the question is ill-posed, because the expression really ought to be understood as
$$\Pi(N(e^{-a^\dagger a}))=|0\rangle\langle0|,$$
meaning that while $e^{-a^\dagger a}$ is still an element of the free algebra and it therefore makes perfect sense to apply the normal ordering to it, to then identify the resulting algebra element to $|0\rangle\langle0|$ we're implicitly applying the CCRs (as of course is pointed out in the answer itself). And after we've done that, it doesn't make sense to ask about normal ordering anymore, because normal ordering is simply not defined on these objects.
