$\mathcal{A}_1 \cup \mathcal{A}_2$ is a $\sigma$-algebra $\iff$ $\mathcal{A}_1 \subseteq \mathcal{A}_2$

Suppose $A_1 \in \mathcal A_1 \backslash \mathcal A_2$ and $A_2 \in \mathcal A_2 \backslash \mathcal A_1$. Then also $A_1^c \in \mathcal A_1 \backslash \mathcal A_2$ and $A_2^c \in \mathcal A_2 \backslash A_1$.

$A_1 \cap A_2$ must be in at least one of $\mathcal A_1$ and $\mathcal A_2$: wlog suppose it is in $\mathcal A_1$. Then $A_2 \backslash A_1 $ can't be in $\mathcal A_1$ because otherwise $A_2 = (A_1 \cap A_2) \cup (A_2 \backslash A_1)$ would be in $\mathcal A_1$. So $A_2 \backslash A_1$ must be in $\mathcal A_2$, and then $A_1 \cap A_2 = A_2 \backslash (A_2 \backslash A_1) \in \mathcal A_2$. Thus $A_1 \cap A_2 \in \mathcal A_1 \cap \mathcal A_2$.

But now consider $A_1^c \cap A_2^c$. This can't be in $\mathcal A_1$, else $A_2^c = (A_1^c \cap A_2^c) \cup (A_1 \backslash (A_1 \cap A_2)) \in \mathcal A_1$, and it can't be in $\mathcal A_2$, else $A_1^c = (A_1^c \cap A_2^c) \cup (A_2 \backslash A_1) \in \mathcal A_2$. So we have a contradiction.

Let $A_1\in\mathcal A_1\setminus\mathcal A_2$ and $A_2\in\mathcal A_2\setminus\mathcal A_1$. Suppose that $\mathcal A_1\cup\mathcal A_2$ is a $\sigma$-algebra. Then every combination of $A_1$ and $A_2$ is contained in $\mathcal A_1\cup\mathcal A_2$, and hence in one of those two.

WLOG, suppose that $A_1\cup A_2\in\mathcal A_1$. Then $A_1\cap A_2\in\mathcal A_2$, since $A_2 = (A_1\cup A_2)\setminus(A_1\cap A_2)$. And since $A_1 = (A_1\setminus A_2)\cup(A_1\cap A_2)$, you have $A_1\setminus A_2\in\mathcal A_1$. But then $$ A_2 = (A_1\cup A_2)\setminus(A_1\setminus A_2)\in \mathcal A_1, $$ a contradiction!