Would duplicates matter in cartesian product of a set? For example:
\begin{align}
A &= \{1, 1, 2\}
\\
B &= \{3, 3, 3, 2, 2, 4\}
\end{align}
Would $A$ cross $B$ equate to $\{(1,3),(1,2),(1,4),(2,3),(2,2),(2,4)\}$ without the dupes of $(1,3)$, etc.

No, because the Cartesian product of sets is itself a set. For sets in general, we consider a set, and a set with the same entries but some duplicates, to be precisely the same.
For example, let $A=\{1,2\},B=\{3,4\},A'=\{1,1,2,2\},B'=\{3,3,4,4,4,4,4\}$. Under these conditions, $A=A',B=B',$ and in turn $A \times B = A' \times B'$.
Thus, the inclusion of duplicate elements does not matter here. Why you would want to include them is beyond me, unless you're working with objects like multisets (which do admit duplicate elements). In such an instance, obviously the opposite holds true. But unless stated otherwise I imagine the first instance holds for you, i.e. you should be removing duplicates from sets.
