Answer to your second question:
These sets will belong to the product $\sigma$-algebra, but not every set in it will be such a set.
Realize that the complement of e.g. $\{1,2\}\times\{1,3,5\}$ is not a set that can be written as $A\times B$.
As you know $\sigma$-algebras are closed under complementation.
Answer to your first question:
In this special case there are partitions $\mathcal P=\{\{1,2\},\{3,4,5\},\{6\}\}$ and $\mathcal Q:=\{\{1,3,5\},\{2,4,6\}\}$ on set $E$. A set is element of the $\sigma(C)$ if and only if it can be written as a finite union of elements of $\mathcal P$. This union is also allowed to be empty. Likewise a set is element of the $\sigma(Y)$ if and only if it can be written as a finite union of elements of $\mathcal Q$.
This partition induces a partition $\mathcal R:=\mathcal P\times\mathcal Q=\{A\times B\mid A\in\mathcal P,B\in\mathcal Q\}$ on set $E\times E$, and a set is element of $\sigma(C)\otimes\sigma(Y)$ if it can be written as a finite union of elements of $\mathcal R$.
Note: because set $E$ is finite the class of algebras on it coincides with the class of $\sigma$-algebras on it.
By definition, $\mathcal{A} \otimes \mathcal{B}$ is the $\sigma$-algebra on $X \times Y$ generated by the set $\mathcal{A} \times \mathcal{B}=\{A \times B: A \in \mathcal{A}, B \in \mathcal{B}\}$, (or equivalently the smallest $\sigma$-algebra that makes the projections measurable).
As $\mathcal{E} \subseteq \sigma_X(\mathcal{E}) = \mathcal{A}$ by definition and likewise $\mathcal{F} \subseteq \sigma_Y(\mathcal{F}) = \mathcal{B}$, we have that
$$\mathcal{E} \times \mathcal{F} = \{E \times F: E \in \mathcal{E}, F \in \mathcal{F}\} \subseteq \mathcal{A} \times \mathcal{B}$$
so $$\mathcal{C} = \sigma_{X \times Y}(\mathcal{E} \times \mathcal{F}) \subseteq \sigma_{X \times Y}(\mathcal{A} \times \mathcal{B}) = \mathcal{A} \otimes \mathcal{B}$$
This uses the obvious fact (by the definitions) that if $\mathcal{G},\mathcal{G}'$ are families of subsets of any set $Z$, then $\mathcal{G} \subseteq \mathcal{G}'$ implies $\sigma_Z(\mathcal{G}) \subseteq \sigma_Z(\mathcal{G}')$ as well.
The simple example in this post shows that we indeed need some condition like $X \in \mathcal{E}$ and $Y \in \mathcal{F}$ to show the reverse inclusion $\mathcal{A} \otimes \mathcal{B} \subseteq \mathcal{C}$ as well. For this inclusion it suffices to show that $$\mathcal{A} \times \mathcal{B} \subseteq \sigma_{X \times Y}(\mathcal{E} \times \mathcal{F}) = \mathcal{C}\tag{1}$$ and this is more subtle:
Define $$\mathcal{A}' = \{A \subseteq X: (\pi_X)^{-1}[A] \in \mathcal{C}\}$$
where $\pi_X: X \times Y \to X$ is the projection.
It is easy to check that this defines a $\sigma$-algebra on $X$, by the properties of inverse images and the fact that $\mathcal{C}$ is a $\sigma$-algebra. Also, for $E \in \mathcal{E}$ (and because $Y \in \mathcal{F}$), we have that
$$(\pi_X)^{-1}[E] = E \times Y \in \mathcal{E} \times \mathcal{F} \subseteq \mathcal{C}$$
so that $\mathcal{E} \subseteq \mathcal{A}'$ which means that
$\sigma_X(\mathcal{E}) = \mathcal{A} \subseteq \mathcal{A}'$ as well, or equivalently:
$$\forall A \in \mathcal{A}: A \times Y \in \mathcal{C}\tag{2}$$
Using the analogous argument for $\mathcal{B}$ and $\pi_Y$ and the assumption that $X \in \mathcal{E}$ we also get:
$$\forall B \in \mathcal{B}: X \times B \in \mathcal{C}\tag{3}$$
And then note that $(2)$ together with $(3)$ imply $(1)$ by the simple fact that
$$A \times B = (X \times B) \cap (A \times Y)$$
using that $\mathcal{C}$ is closed under intersections.
This concludes the proof of the reverse inclusion.
Best Answer
Of course if you take $A\times\emptyset$ or $\emptyset\times B$, they're equal to $\emptyset$ which is in the product $\sigma$-algebra $\pmb{F}\times \pmb G$even if $A\notin\pmb F$ or $B\notin\pmb G$.
However, assume $A\times B\in\pmb F\times \pmb G$ such that $A\neq \emptyset$ and $B\neq\emptyset$. Then we know the following:
So take $x\in A$ (and that's where we use $A\neq\emptyset$). The $x$-section of $A\times B$ is $B$. According to the proposition above, $B\in\pmb G$, and you can similarly find that $A\in\pmb F$.