Question: If $\emptyset\neq\{\emptyset\}$, then the result of $\mathbf{A}\times\emptyset$ should differ from the result of $\mathbf{A}\times\{\emptyset\}$. 

Does it? What is the result of the latter, if - let's say - $\mathbf{A}=\{1, 2\}$?

Answer: 
Yup.

The Cartesian product of $X$ and $Y$ is the set of pairs $(a, b)$ with $a\in X$ and $b\in Y$.

The Cartesian product $A\times\emptyset$ is empty because there are no pairs of that form - $\emptyset$ has no elements!

But $\{\emptyset\}$ **does** have an element - namely, $\emptyset$. For example, if we take $A=\{1, 2\}$, then $$A\times\{\emptyset\}=\{(1, \emptyset), (2, \emptyset)\}.$$ Remember that the emptyset is *not nothing*! It just *contains* nothing. Think about it as an empty bag - it's still a thing that exists (namely, a bag).
Recall that $X\times Y$ is the set of all pairs $(x,y)$ such that $x\in X$ and $y\in Y$. If $Y=\emptyset$ then no such pairs satisfy this condition, so $A\times\emptyset=\emptyset$. On the other hand, the set $\{\emptyset\}$ has precisely one element, and so $A\times\{\emptyset\}=\{(a,\emptyset)\,:\,a\in A\}$ which is nonempty if $A$ is nonempty, and in fact there is an obvious bijection $A\longrightarrow A\times\{\emptyset\}$.