I want to show that $\sigma\{A\times B: A\in \sigma(\mathcal{E_1}),B\in \sigma(\mathcal{E_2})\} = \sigma\{E_1\times E_2: E_1\in \mathcal{E_1},E_2\in \mathcal{E_2}\}$.
I see that the generating set $\{E_1\times E_2: E_1\in \mathcal{E_1},E_2\in \mathcal{E_2}\}\subseteq \{A\times B: A\in \sigma(\mathcal{E_1}),B\in \sigma(\mathcal{E_2})\}$. Yet for the other direction, I tried to demonstrate that for any $A\times B$ such that$ A\in \sigma(\mathcal{E_1}),B\in \sigma(\mathcal{E_2}) $ then $A\times B \in \sigma\{E_1\times E_2: E_1\in \mathcal{E_1},E_2\in \mathcal{E_2}\}$. But failed to make it.
I am going to suppose you are working with on the product of the set $X$ and $Y$, i.e. your sigma-algebras are defined on $X\times Y$. Now, denote $$ \mathcal{E} = \{E_1\times E_2: E_1\in \mathcal{E_1},E_2\in \mathcal{E_2}\} $$ You can show that $$ \{A\subseteq X : A\times Y \in \sigma(\mathcal{E})\} $$ is a $\sigma$-algebra on $X$ that contains $\mathcal{E_1}$, and thus $\sigma(\mathcal{E_1})$. Likewise, $$ \{B\subseteq Y : X\times B \in \sigma(\mathcal{E})\} $$ is a $\sigma$-algebra on $Y$ that contains $\mathcal{E_2}$, and thus $\sigma(\mathcal{E_2})$.
In particular, $A\times Y \in \sigma(\mathcal{E})$ for all $A\in \sigma(\mathcal{E_1})$ and $X\times B \in \sigma(\mathcal{E})$ for all $B\in \sigma(\mathcal{E_2})$. Ergo, $$ A\times Y \cap X\times B = A\times B \in \sigma(\mathcal{E}) \quad \text{whenever } A\in \sigma(\mathcal{E_1}), B\in \sigma(\mathcal{E_2}) $$ We conclude that $$ \sigma\{A\times B: A\in \sigma(\mathcal{E_1}),B\in \sigma(\mathcal{E_2})\} \subseteq \sigma(\mathcal{E})= \sigma\{E_1\times E_2: E_1\in \mathcal{E_1},E_2\in \mathcal{E_2}\} $$