An indexed family of sets is a function $A=(i\mapsto A_i)_{i\in I}:I\to E$ for some $I\ne \emptyset$ and some $E$, where each $A_i$ is a set.
I suppose $E$ is a set that contains all $A_i$ and possibly other sets or other atoms (correct me if I'm wrong).
Why doesn't one define an indexed family of sets as the set consisting of all sets $A_i$ where $i$ ranges over $I$? Isn't that what our intuition invites us to do? After all, it is not uncommon to say something like "consider a set $A'$ from the family of sets $\{A_i: i\in I\}$ ", and noone ever assumes that $A'$ is actually a function; one just assumes $A'=A_{i_0}$ for some $i_0\in I$.
And here is a related concern:
The product of an indexed family of sets $A$ as above is $$\prod_{i\in I} A_i=\{f:I\to \bigcup_{i\in I} A_i: (\forall i\in I)(f(i)\in A_i)\}$$
Is it not the same as the Cartesian product (at least for $I$ finite)? (This definition does not agree with the standard definition when $I=\{1,2\}$ in which case $A_1\times A_2=\{(a_1,a_2):a_i\in A_i \text{ for } i=1,2\}$.) Also, what "bad" happens if we don't require $(\forall i\in I)(f(i)\in A_i)$? E.g. if $f(2)\in A_1$.
Consider the indexed family $(A_i:i\in\{1,2\})$ where $A_1=\{5,6,7\}$ and $A_2=\{3,8\}$. Consider a second indexed family $(B_i:i\in\{1,2\})$ where $B_1=\{3,8\}$ and $B_2=\{5,6,7\}$. And consider a third indexed family $(C_i:i\in\{7,2\})$ where $C_7=\{3,8\}$ and $C_2=\{5,6,7\}$. If I were to identify, as you suggested, these families with $\{A_i:i\in\{1,2\})\}$, $\{B_i:i\in\{1,2\}\}$, and $\{C_i:i\in\{7,2\}\}$, then they would all be the same. I wouldn't really have indexed families at all, because the indexing has been lost.
Concerning products, let's use the first example above. Work out what the definition gives for $\prod_{i\in\{1,2\}}A_i$. You'll get a set of $6$ functions. As you already pointed out in the question, that set is not the same as $A_1\times A_2$ --- but why should it be? It has an obvious one-to-one correspondence with $A_1\times A_2$, and that's all that one needs. Although the definition of $A_1\times A_2$ can be extended (in more than one way) to define products of $3$ or more factors, that seems to work only as long as the number of factors is finite. The $\prod$ definition, in contrast, makes sense even for infinite $I$.
Finally, if you omit, from the definition of $\prod$, the requirement that $f(i)\in A_i$, then, in that same first example above, you'd get that $\prod_{i\in\{1,2\}}A_i$ has not $6$ but $25$ elements. Furthermore, you'd get the same product if $A_1$ and $A_2$ were $\{5,8\}$ and $\{3,6,7\}$, respectively. Not how a product should behave.