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