If $a\colon A_1\to A_2$ and $b\colon B_1\to B_2$ are isomorphisms, then $c=a\times b\colon A_1\times B_1\to A_2\times B_2$ is also an isomorphism.
You already showed it’s a morphism (the argument is essentially correct, but you should start from $c\bigl( (x_1,y_1)(x_2,y_2)\bigr)$, not from $c(x_1x_2,y_1y_2)$).
(Also, you should use notation that is suggestive! If your domain is called $A_1\times B_1$, then why not refer to the elements of $A_1$ by lower case $a$-s and those of $B_1$ by lower case $b$-s?)
Anyhow... two different ways of showing it is an isomorphism:
It has an inverse: we know $a$ has an inverse $a^{-1}\colon A_2\to A_1$ and that $b$ has an inverse $b^{-1}\colon B_2\to B_1$. Show that $d=a^{-1}\times b^{-1}\colon A_2\times B_2\to A_1\times B_1$ is the inverse of $c$.
Show $c$ is bijective. To show that $c$ is one-to-one, consider the kernel: if $c(x,y) = (e,e)$, show that then $a(x)=e$ and $b(y)=e$, and use the fact that $a$ and $b$ are both one-to-one to conclude $c$ is one-to-one. And to show $c$ is surjective, take $(z,w)\in A_2\times B_2$; use the fact that $a$ and $b$ are both surjective to show that you can find an element in $A_1\times B_1$ whose image under $c$ is $(z,w)$.
Pick your favorite method.
No.
The first condition can’t be satisfied by any $x_1, \dotsc, x_n$ that has two size-$\lfloor n/2\rfloor$ subsets $A, B$ with the same sum, because one can take $x_1, \dotsc, x_n$, remove the elements of $A$, insert the elements of $B$ (in sorted order), and get $y_1, \dotsc, y_n$ such that $\sum x_i = \sum y_i$. (For example, $x_1, …, x_7 = 1, 2, 3, 10, 16, 19, 22$ doesn’t work because $1 + 10 + 16 = 2 + 3 + 22$ and we can take $y_1, …, y_7 = 2, 2, 3, 3, 19, 22, 22$.)
Therefore, the $\binom{n}{\lfloor n/2\rfloor}$ sums of the size-$\lfloor n/2\rfloor$ subsets of $\{x_1, \dotsc, x_n\}$ must all be distinct integers. So at least one of these sums must have absolute value at least $\frac12\left(\binom{n}{\lfloor n/2\rfloor} - 1\right)$, which means one of the $|x_i|$ must be at least $\frac{1}{2\lfloor n/2\rfloor}\left(\binom{n}{\lfloor n/2\rfloor} - 1\right) = Θ(2^n n^{-3/2})$, which is exponential in $n$.
Best Answer
Counterexample:
$12+4+3 \ =\ 9+8+2$
$12\cdot4\cdot3 \ = \ 9\cdot8\cdot2$
Moreover, for $\ i>2\ ,\ $you can always find infinitely many counterexamples.