$\sigma$- ideal 

Let $(\Omega,\mathcal{A})$ be a measurable space. $\mathcal{N}\subset\mathcal{P}(\Omega)$ is called a $\sigma$ ideal, if
    $$
(1)~\emptyset\in\mathcal{N},~~~~~(2) N\in\mathcal{N}, M\subset N\Rightarrow M\in\mathcal{N},~~~~~(3)(N_n)\in\mathcal{N}^{\mathbb{N}}\Rightarrow\bigcup_n N_n\in\mathcal{N}
$$
    Show that for every $\sigma$-ideal $\mathcal{N}$ it is
    $$
\sigma(\mathcal{A}\cup\mathcal{N})=\left\{A\Delta N|A\in\mathcal{A},N\in\mathcal{N}\right\}.
$$
    Hint: 
    $$
\left\{A\Delta N|A\in\mathcal{A},N\in\mathcal{N}\right\}=\left\{B\subset\Omega|\exists A\in\mathcal{A},N\in\mathcal{N}: B\setminus N=A\setminus N\right\}
$$



I do not have a special idea, to be honest.
For the inclusion "$\subseteq$ " I thought that maybe this is a strategy:
(1) Show that $\mathcal{S}:=\left\{A\Delta N|A\in\mathcal{A},N\in\mathcal{N}\right\}$ is a $\sigma$-algebra.
(2) Show that $\mathcal{A}\cup\mathcal{N}\in\mathcal{S}$.
Is that right or helpful? If yes: Which is the strategy for the other inclusion?
 A: Yes, for $\subseteq$ this would be enough. (2) holds trivially, as $\emptyset\in\mathcal A\cap\mathcal N$, so that $A=A\Delta \emptyset\in\mathcal S$ and $N=\emptyset\Delta N\in\mathcal S$. 
For (1), observe that $A\Delta N=A\cup N'$ where $N'=N\setminus A \subseteq N$ so $N'\in\mathcal N$. Thus, we have
$$\bigcup_n(A_n\Delta N_n)=\bigcup_n(A_n\cup N_n') =\bigcup_nA_n\,\cup\,\bigcup_nN_n' =\bar A\cup\bar N\,,$$
where $\bar A=\bigcup_n A_n\in\mathcal A$ and $\bar N=\bigcup_n N_n'\in\mathcal N$.
Applying the same trick backwards, let $\bar N':=\bar N\setminus\bar A$, it is still in the $\sigma$-ideal $\mathcal N$. As $\bar N'$ and $\bar A$ are disjoint, we have
$$\bar A\cup\bar N=\bar A\cup\bar N'=\bar A\Delta\bar N'\,.$$
Similarly, $(A\Delta N)^\complement = (A\cup N')^\complement=A^\complement\cap (N')^\complement=A^\complement\setminus N'=A^\complement\Delta N''\ $ with $N'':=A^\complement\cap N'\ \in\mathcal N$.
And, for the other inclusion,
(3) Show that whenever a $\sigma$-algebra $\mathcal U$ contains $\mathcal A\cup\mathcal N$, it also contains $\mathcal S$.
A: Your strategy is a good one.

For the opposite direction, you want to show:

If $\Sigma$ is a $\sigma$-algebra, then $\mathcal A \cup \mathcal N$ implies $\mathcal S \subseteq \Sigma$.

(which is precisely saying that $\mathcal S$ fulfils the definition of generated $\sigma$-algebra).
In this case, I suggest to fix an element $A \Delta N \in \mathcal S$, and try to construct it from $\mathcal A \cup \mathcal N$ using the closure rules for a $\sigma$-algebra. (E.g. if $A_1,\ldots A_n \in \mathcal A$, then we know that $\bigcup_n A_n \in \Sigma$ for any $\sigma$-algebra $\Sigma$ containing $\mathcal A$ -- this is the closure rule for "countable union"). 
Hint:

 $A \setminus N = A \cap N^\complement = (A^\complement \cup N)^\complement$


A hint for the union part of the initial direction (I assume you have either proved the hint given in the exercise or are allowed to use it without proof):

 \begin{align}\bigcup_n B_n \setminus \bigcup_m N_m&= \bigcup_n \left(B_n \setminus \left(\bigcup_m N_m\right)\right)\\&= \bigcup_n\left(\left(B_n \setminus N_n\right)\setminus \bigcup_{m \ne n} N_m\right)\\&= \bigcup_n \left(\left(A_n \setminus N_n\right)\setminus \bigcup_{m \ne n} N_m\right)\\&= \bigcup_n A_n \setminus \bigcup_m N_m\end{align}

I'll leave complement and containment of empty set to you.
