Given any family of sets $\mathcal{F}$ is it true that:
$$\bigcup_{S\in \mathcal{F}} \left(S\times S\right)=\left(\bigcup_{S\in \mathcal{F}} S\right)\times \left(\bigcup_{S\in \mathcal{F}} S\right)\implies \left(\bigcup_{S\in \mathcal{F}} S\right)\in \mathcal{F}$$
The converse I can tell is clearly true, however I'm not totally sure about the above.
I might just be not thinking as its early in the morning, and this could be trivially wrong/right.
No, it is not. Take $\mathcal{F} = \Big\{ S_1=\{1,2\},S_2=\{2,3\},S_3=\{3,1\}\Big\}$ and let $$X :=\bigcup_{S\in \mathcal{F}} S = \{1,2,3\}$$ then $X\notin \mathcal{F}$. But we have
$$\bigcup_{S\in \mathcal{F}} \left(S\times S\right) = \{(1,1),(1,2),(2,1),(2,2),(1,3),(3,1),(3,3),(2,3),(3,2),(2,2)\} =X\times X$$