Your proof is in the right direction. Here I copied it and I changed some points in it to make it a complete proof.
Theorem 1.9 - Suppose that $(X,M,\mu)$ is a measure space. Let $\mathcal{N} = \{N\in M:\mu(N) = 0\}$ and $\overline{M} = \{E\cup F: E\in M, F\subset N, N\in\mathcal{N}\}$. Then $\overline{M}$ is a $\sigma$-algebra and there is a unique extension $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$.
Claim 1 - $\overline{M}$ is a $\sigma$-algebra
proof:
i.) Since $M$ is a $\sigma$-algebra, $\emptyset\in M\subset \overline{M}$ so $\emptyset\in \overline{M}$.
ii.) Suppose $B\in\overline{M}$ then there is an $E\in M$ such that $B = E\cup F$ where $F\subset N$ and $N\in\mathcal{N}$. Then
\begin{align*}
X\setminus B &= X\cap B^c\\
&= X\cap (E\cup F)^c\\
&= X\cap ((E^c\cap N^c\cap F^c)\cup(F^c\setminus N^c))\\
&= X\cap ((E^c\cap N^c)\cup (F^c\cap N))\\
&= X\cap ((E\cup N)^c\cup (N\setminus F))
\end{align*}
Since $(E\cup N)^c\in M$ and $(N\setminus F)\subset N\in M$ with $\mu(N) = 0$, then $B^c\in\overline{M}$.
iii.) Let $\{B_j\}_{1}^{\infty}\in\overline{M}$ then for each $j$ there is an $E\in M$ such that $B_j = E_j\cup F_j$ where $F_j\subset N_j$ and $\mu(N_j) = 0$. So, $$\bigcup_{1}^{\infty}B_j = \bigcup_{1}^{\infty}(E_j\cup F_j) = \bigcup_{1}^{\infty}E_j \cup \bigcup_{1}^{\infty}F_j$$ Note that $\bigcup_{1}^{\infty}F_j\subset \bigcup_{1}^{\infty}N_j$ and $\mu\left(\bigcup_{1}^{\infty}N_j\right) = 0$. So we have $\bigcup_{1}^{\infty}B_j\in\overline{M}$. Therefore $\overline{M}$ is a $\sigma$-algebra.
Claim 2 - There is a unique extention $\overline{\mu}$ of $\mu$ to a complete measure on $\overline{M}$
Proof:
We first need to show that $\overline{\mu}$ is well-defined. Suppose $E\cup F\in\overline{M}$, set $\overline{\mu}(E\cup F) = \mu(E)$. This is well-defined since if $E_1\cup F_1 = E_2\cup F_2$ where $F_j\subset N_j\in\mathcal{N}$. Then we know that $E_1\subset E_1\cup F_1 = E_2\cup F_2 = E_2\cup N_2$ then $E_1\subset E_2\cup N_2$ and so by monotonicity $\mu(E_1)\leq \mu(E_2) + \mu(N_2) = \mu(E_2)$. Also, since $E_2\subset E_2\cup F_2 = E_1\cup F_1 = E_1\cup N_1$ then again by monotonicity $\mu(E_2)\leq \mu(E_1)+\mu(N_1) = \mu(E_1)$. So we have $\mu(E_1) = \mu(E_2)$. Thus $\overline{\mu}$ is well-defined.
I now need to show that $\overline{\mu}$ is a complete measure on $\overline{M}$ and that $\overline{\mu}$ is the only measure on $\overline{M}$ that extends $\mu$.
Step 1 - Show $\overline{\mu}$ is a measure.
Proof:
i.) $\overline{\mu}(\emptyset) = \overline{\mu}(\emptyset \cup \emptyset)= \mu(\emptyset) = 0$ (here we took $E=\emptyset$ and $F= N=\emptyset$).
ii.) Let $\{A_n\}_{1}^{\infty}\in\overline{M}$, disjoint, then there is an $\{E_n\}_{1}^{\infty}\in M$ and a sequence $\{F_n\}_{1}^{\infty}\subset N\in\mathcal{N}$ such that $A_n = E_n\cup F_n$ for all $n$. Note that $\{E_n\}_{1}^{\infty}$ is a family of disjoint sets in $M$. Note also from part (iii.) of claim 1, $\cup_{1}^{\infty}F_n\subset \cup_{1}^{\infty}N_n$ and $\mu(\cup_{1}^{\infty}N_n)=0$. Thus,
\begin{align*}
\overline{\mu}\left(\bigcup_{1}^{\infty}A_n\right) = \overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup F_n\right) &=
\overline{\mu}\left(\bigcup_{1}^{\infty}E_n\cup \bigcup_{1}^{\infty}F_n\right)\\
&= \mu\left(\bigcup_{1}^{\infty}E_n\right)=\\
&= \sum_{1}^{\infty}\mu(E_n)=\\
&= \sum_{1}^{\infty}\overline{\mu}(E_n\cup F_n) \\
&= \sum_{1}^{\infty}\overline{\mu}(A_n)
\end{align*}
Thus $\overline{\mu}$ is a measure.
Step 2 - Show $\overline{\mu}$ is a complete measure.
Proof:
Let $A\subset X$ and suppose there is an $B\in\overline{M}$ such that $A\subset B$ and $\overline{\mu}(B) = 0$. Since $B\in\overline{M}$, we know we can write
$B = E\cup F$ where $E\in M$ and $F\subset N\in M$ with $\mu(N) = 0$. Since $\overline{\mu}(B) = 0$ and $\overline{\mu}(B) = \mu(E)$, we have $\mu(E)=0$. So
$E\cup N \in M$ and $\mu(E\cup N)=0$.
Now note that $A=\emptyset \cup A$, $\emptyset \in M$ and $A\subseteq B = E\cup F \subseteq E\cup N$ and $E\cup N \in M$ and $\mu(E\cup N)=0$. So $A \in \overline{M}$ and $\overline{\mu}(A)=\mu(\emptyset)=0$. So $\overline{\mu}$ is complete.
Step 3: Show $\overline{\mu}$ is a unique extension of $\mu$
3.1. $\overline{\mu}$ is an extension of $\mu$.
Given $A \in M$, we have that $A=A\cup\emptyset$ so $A=A\cup F$ where $F\subset N$, $N\in M$ and $\mu(N)=0$ (just take $F=N=\emptyset$). Thus, $A \in \overline{M}$ and
$\overline{\mu}(A) =\overline{\mu}(A\cup \emptyset)=\mu(A)$.
So we proved that $M \subseteq \overline{M}$ and $\overline{\mu}$ is an extension of $\mu$.
3.2. Uniqueness.
Suppose $\nu$ is an extension of $\mu$ to $\overline{M}$. Let $A\in\overline{M}$, then there is $E\in M$, $N\in M$ with $\mu(N)=0$ and $F\subseteq N$ such that $A=E\cup F$. Then, since $E\in M$ and $E\cup N\in M$,
$$\mu(E)=\nu(E)\leqslant \nu(A)=\nu(E\cup F)\leqslant \nu(E\cup N)=\mu(E\cup N)
\leqslant \mu(E)+\mu(N)=\mu(E)$$
So $\nu(A)=\mu(E)=\overline{\mu}(A)$. So $\nu = \overline{\mu}$.
Best Answer
The problem is that $0$ is not the only extended real number $a$ such that $a+a=a$, since $\infty+\infty=\infty$. So you could also have $\mu(\emptyset)=\infty$ instead of $\mu(\emptyset)=0$. (In that case, you would have $\mu(A)=\infty$ for all $A$ since you can write $A=A\cup\emptyset$, so as long as any set has finite measure this can't happen.)