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 $A\longrightarrow A\times\{\emptyset\}$.
The number of elements in $\varnothing$ is $0$. 

The number of elements in $\{\varnothing\}$ is $1$. 

For the cartesian product, the number of elements is the product. So, if $A=\{1,2\}$ has two elements, we see:

$A \times \varnothing$ has $2 \times 0 = 0$ elements, but 

$A \times \{\varnothing\}$ has $2 \times 1 = 2$ elements.