I have the measure space $(X,\mathbb{E},\mu)$ where $\mathbb{E}$ is a $\sigma$-algebra on $X$ and $\mu$ is the Lebesgue measure.
I have that $$\mathbb{N}_\mu=\{E\subseteq X\; |\; \text{there exist }E\in\mathbb{E}\;\text{so}\;N\subseteq E\;\;\text{and}\;\mu(E)=0\}\in\mathbb{N}_\mu\},$$ so is the following argumentation valid:
The enlargement of $\mathbb{E}$ with respect to $\mu$ is: $$\mathbb{E}_\mu=\mathbb{E}\cup\mathbb{N}_\mu,$$ so $$\mathbb{E}\subseteq \mathbb{E}_\mu\quad\quad\text{and}\quad\quad\mathbb{N}_\mu\subseteq \mathbb{E}_\mu$$ as $\mathbb{E}=\mathbb{E}_\mu$ \ $\{\mathbb{N}_\mu\}$ and $\mathbb{N}_\mu=\mathbb{E}_\mu$ \ $\{\mathbb{E}\}$.
If you start with $(X,\mathcal{M},\mu)$, the completed $\sigma$-algebra is
$$\mathcal{M}^*=\{ B \subseteq X : \text{ there exists } A,C \in \mathcal{M} \: \mu(C)=0 \wedge B \setminus A \subseteq C \}.$$
Note that fixing $C=\emptyset$ inside that set builder reproduces $\mathcal{M}$ itself, so no union is required. Similarly fixing $A=\emptyset$ reproduces all of the actual null sets in the completed space.
This is not quite what you wrote, what you wrote has some minor errors that make it rather nonsensical (for example $E$ being bound under two different quantifiers in the same expression).