Which of the following subsets of $\mathbb R^2$ with the indicated operation, is a group? Which is an abelian group? 
$(a, b) * (c, d) = (ac, bc + d),$ on the set $\{(x, y) \in \mathbb R \times \mathbb R:  x \neq 0\}$.

$(a, b) * (c, d) = (ac, bc + d) = (ca, da + b) = (c, d) * (a, b).$ Elements here commute about $*$.
$((a, b) * (c, d)) * (e, f) = ((ac, bc + d)) * (e, f) = (ace, bce + de + f) \neq (eac, fac + bc + d) = (e, f) * ((a, b) * (c, d)).$ Associativity fails.
$(a, b) * (e_1, e_2) = (ae_1, be_1 + e_2) = (a, b),$ so $(e_1, e_2) = (1, 0).$ Thus, $(e_1, e_2) * (a, b) = (a, b).$ Identity exists.
$(a, b) * (a', b') = (aa', ba' + b') = (e_1, e_2),$ $(a', b') = (\frac {1}{a}, - \frac {b}{a})$. Thus, $(a', b') * (a, b) = (1, 0).$ Inverse exists.
This set is not a group.

Same operation as above, but on the set $\mathbb R \times \mathbb R.$

In this case inverse will fail since we can't divide by $0.$

$(a, b) * (c, d) = (ac - bd, ad + bc),$ on the set $\mathbb R \times \mathbb R$ with the origin deleted.

$(a, b) * (c, d) = (ac - bd, ad + bc) = (ca - db, cb + da) = (c, d) * (a, b)$. Commutativity holds.
$((a, b) * (c, d)) * (e, f) = ((ac - bd, ad + bc)) * (e, f) = (ace - bde - (adf + bcf), acf - bdf + ade + bce =  eac - ebd - (fad + fbc), ead + ebc + (fac - fbd) = (e, f) * ((a, b) * (c, d)).$ Associativity holds.
$(a, b) * (e_1, e_2) = (ae_1 - be_2, ae_2 + be_1) = (a, b).$ So, $ e_1 = \frac {a + be_2}{a}, e_2 = \frac {b - be_1}{a}.$
$(e_1, e_2) * (a, b)$  produces a terrible mess that doesn't equal $(a, b).$ No identity and no inverse.

Consider the operation of the preceding problem on the set $ \mathbb R^2$. Is this a
  group? Explain.

Assuming the preceding problem is done correctly, this one won't be a group either because $2$ axioms don't hold. 
Need my work checked.
 A: Let $$S_0=\{(x,y)∈R×R:x≠0\}$$ and define $*$ on $S$ such that
$$(a,b)∗(c,d)=(ac,bc+d),\,\,\,\,\ ac\not=0.$$
Since $$(a,b)∗(c,d)\in S_0,\,\,\,\,\forall(a,b),(c,d)\in S_0$$
this is a binary operation on $S_0$ ($S_0$ is closed under multiplication).
Note that $$(a,b)∗(c,d)=(ac,bc+d)$$
$$(c,d)*(a,b)=(ca,da+b)$$
$$((a,b)∗(c,d))*(e,f)=(ac,bc+d)*(e,f)=(ace,ebc+ed+f)$$
$$(a,b)∗((c,d)*(e,f))=(a,b)*(ce,de+f)=(ace,bce+de+f).$$
Also $$(a,b)*(1,0)=(a,b)$$
$$(a,b)*\Big(\frac{1}{a},-\frac{b}{a}\Big)=(1,0).$$ Therefore $$(a,b)∗(c,d)\not=(c,d)*(a,b),\,\,\,\,((a,b)∗(c,d))*(e,f)=(a,b)∗((c,d)*(e,f))$$ and also $(1,0)\in S_0.$

Note that, Associativity holds and Commutativity fails. 

Hence $(S_0,*)$ form a non-abelian group.
If you define the same operation $*$ on the set $S=R×R,$ it will not be a group as elements of the form $(0,x)$ have no inverses.
A: (a,b)∗(c,d)=(ac,bc+d),(a,b)∗(c,d)=(ac,bc+d), on the set {(x,y)∈R×R:x≠0}{(x,y)∈R×R:x≠0}.
This is not commutative--as apparent in the OP's own math! Therefore the OP fails to find it associative, which it is, but not if you commute first.  
Anyone else working through this book feel free to drop me a line. 
