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).
Yes, it does differ. Say, for your example, $\{1,2\}\times\{\varnothing\}=\{(1,\varnothing),(2,\varnothing)\}$.