A partial answer:
Let $\psi \in \mathcal D(N)$; in particular, it follows that $\psi \in \mathcal D(N^{1/2})$. We want to show that these conditions are sufficient such that e.g. $a(f)\psi \in \mathcal D(N^{1/2})$. Then the product $a^\dagger(g) a(f) \psi$ would be well-defined.
To do so, I think we can use the fact that for all $\psi \in \mathcal D(N^{1/2})$ it holds that
$$||a(f)\psi||_F \leq ||f||_{\mathfrak h}\, ||N^{1/2}\psi||_F \quad , \quad .\tag{1}$$
where $\mathfrak h$ denotes the underlying one-particle space and $f\in \mathfrak h$. We then proceed by noting that if $\psi=(\psi_n)_n \in \mathcal D(N)$ we have that
$$ \infty > ||f||_\mathfrak h ||N\psi||_F^2 = ||f||_\mathfrak h\sum\limits_n n^2||\psi_n||^2 = ||f||_\mathfrak h \sum\limits_n n ||(N^{1/2}\psi)_n||^2 \geq \sum\limits_n n ||(a(f)\psi)_n||^2 =||N^{1/2}a(f)\psi||_F\tag{2} $$
and hence $a(f)\psi\in \mathcal D(N^{1/2})$.