$\Bbb R -${-1} = {$x \in \Bbb R | x \neq -1$}
We start by factoring $a+b+ab$ into $(1+a)(1+b)-1$
1) Show (R,$\oplus$) is a group
a) closure
Now to show that $a \oplus b$ is closed, we can start by saying that we know $\Bbb R$ is closed under addition and multiplication. Then we just need to show that for $a,b \in \Bbb R - ${-1}, that $a \oplus b \in \Bbb R -${-1}
Let's use proof by contradiction. So suppose that $a+b+ab=-1$. Then $(1+a)(1+b)-1 = -1$
$(1+a)(1+b)=0$
But then either $a=-1$ or $b=-1$, a contradiction.
So $a \oplus b \in \Bbb R -${-1} shows closure under $\oplus$
b) associative
($a \oplus b$) $\oplus$ c = $(a+b+ab) \oplus c$ = $a+b+ab + c +(a+b+ab)c$
= $a+b+ab+c+ac+bc+abc$ = $a + (b+c+bc)+a(b+c+bc)$
= $a \oplus (b+c+bc)$ = $a \oplus (b \oplus c)$
c) identity element
Let $e=0$. Then $a \oplus e = a+0+a(0) = a$ and $e \oplus a = 0+a+(0)a = a$
d) inverses
Suppose $(1+a)(1+b)-1 = 0$. Then $(1+a)(1+b)=1$ and for any $a$, $a^{-1}$ is such that $(1+b)=\frac{1}{1+a} \to b= \frac {1}{1+a} -1$ which is defined for all $\Bbb R$ except {-1}
*(R,$\oplus$) is abelian
Since addition and multiplication are commutative, then $a+b+ab=b+a+ba$
Hence, $a \oplus b = b \oplus a$
This shows the group is abelian
That's part 1
Best Answer
Since you want to avoid thinking about $\mathbb{Z}_n$ as a quotient ring, write $\mathbb{Z}_n=\{[a]\mid a\in\mathbb{Z}\}$, where $$[a]=\{b\in\mathbb{Z}\mid a\equiv b \mbox{ (mod n) }\}$$ is the congruence class of $a$. Then, addition and multiplication are defined by $$[a]+[b]=[a+b]\;\;\mbox{ and }\;\;[a][b]=[ab].$$ We now have enough to prove that $\mathbb{Z}_n$ is a commutative ring (the ring structure being inherited from $\mathbb{Z}$). For example, associativity of addition can be proved as follows: $$ \begin{align} ([a]+[b])+[c]&=[a+b]+[c]\\&=[(a+b)+c]\\&=[a+(b+c)]\\&=[a]+[b+c]\\&=[a]+([b]+[c]). \end{align} $$ All the other axioms can be proved in the same manner.