I came across this question after reading some materials about the Axiom of Choice:
Let $I$ be any nonempty index set and let $(G_i,\star_i)$ be a group for each $i\in I$. The direct product of the groups $G_i$, $i\in I$ is the set $G=\prod_{i\in I} G_i$ with a binary operation defined as follows: if $\prod a_i$ and $\prod b_i$ are elements of $G$, then $$\bigg(\prod_{i\in I}a_i\bigg)\bigg(\prod_{i\in I}b_i\bigg)=\prod_{i\in I}a_i\star_ib_i$$ i.e., the group operation in the direct product is defined componentwise.
(a) Show that this binary operation is well-defined... (Abstract Algebra: Dummit & Foote, Direct products, Ex. 15)
What do I have to show here? Is this definition not well enough? Do I have to invoke the definition of a binary operation $\star:G\times G\rightarrow G$ and show that $\prod_{i\in I}a_i\star_ib_i\in G$?
You are being asked to show that it is essentially a "unique assignment." Which is to say, if $(\prod a_i)(\prod b_i)$ can only be one thing. However, here there is not a lot to show, since if you assume that two products of elements are different, then they result in different elements in the product, almost by definition.
Here is a similarly worded question for quotients.
Regarding you last statement, no, since your assertion is false. You may want to show that $$\prod_{i \in I} a_ib_i \in \prod_{i \in I}G_i$$ as a set, but this is immediate, since $a_i b_i \in G_i$ since each of these $G_i$ are groups in their own right.