We first show that $G = \mathbb{C}^* = \mathbb{C} − \{0\}$ under complex multiplication forms a group.
Let $z = a + bi$ and $w = c + di$ be both in $\mathbb{C}^∗$. Then we have
$zw = (a + bi)(c + di) = ac − bd + (ad + bc)i$. To show that this element is in $G$, we
compute $$(ac − bd)^2 + (ad + bc)^2 = a^2c^2 − 2abcd + b^2 d^2 + a^2 d^2 + 2abcd + b^2c^2
\\= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2
\\= a^2(c^2 + d^2) + b^2(c^2 + d^2)
\\= (a^2 + b^2)(c^2 + d^2).$$
Now $z, w ∈ G$ imply that a $a^2+b^2 > 0$ and $c^2+d^2 > 0$, whence $(ac−bd)^2+(ad+bc)^2 > 0$ and $zw ∈ G$.
To show associativity, let $z = a + bi, w = c + di$ and $u = e + fi$. Then
$$(zw)u = [(a + bi)(c + di)](e + f i)
= [(ac − bd) + (ad + bc)i](e + f i)
\\= [(ac − bd)e − (ad + bc)f] + [(ac + bd)f + (ad + bc)e]i
\\= (ace − bde − adf − bcf) + (acf − bdf + ade + bce)i
\\= [(a(ce − df) − b(cf + de)] + [a(cf + de) + b(ce − df)]i
\\= (a + bi)[(ce − df) + (cf + de)i]
\\= (a + bi)[(c + di)(e + f i)]
\\= z(wu).$$
- Identity Element and Inverse:
It is not difficult to check that $1 + 0i$ acts as a unit in multiplication and that
$$(a + bi)^{−1} =
\frac{1}{a+bi} =\frac{a−bi}{(a+bi)(a−bi)} = \frac{a−bi}{a^2+b^2} = \frac{a}{a^2+b^2} −
\frac{b}{a^2+b^2} i ∈ G$$
, since
$$\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2 = \left(\frac{a^2+b^2}{(a^2+b^2)^2}\right) = \frac{1}{a^2+b^2} > 0$$
Since all the group axioms hold, $(G,\cdot)$, is a group under multiplication.
Ok, this is a group, now we let see that Circle Group is a Group.
Let $K$ be the set of all complex numbers of unit modulus:
$K={z∈\mathbb{C}:|z|=1}$
Then the circle group $(K,\cdot)$ is an uncountably infinite abelian group under the operation of complex multiplication.
$$
z,w ∈ K
⟹ |z| = 1=|w|
⟹ |zw| = |z||w|
⟹ zw ∈ K $$
So $(S,\cdot)$ is closed.
Associativity, comes from complex multiplication is associative.
Identity : From Complex multiplication identity is one we have that the identity element of $K$ is $1+0i$.
Inverses
We have that $|z|=1⟹\frac{1}{|z|}=\left|\frac{1}{z}\right|=1$.
But $z\cdot\frac{1}{z}=1+0i$.
So the inverse of $z$ is $\frac{1}{z}$.
- Commutative: We have that Complex Multiplication is Commutative
So $K$ is a subgroup of $G$ under complex multiplication.
We can assert that from complex multiplication is commutative it also follows from Subgroup of Abelian Group is Abelian that $K$ is an abelian group.
Primes are naturally* thought of as the free generators of a group (the multiplicative group of invertible rational numbers $\Bbb Q^\times$ modulo torsion, i.e. up to sign), not themselves a group. Indeed the freeness aspect means they're as far away as possible from being closed under the operation of multiplication. More precisely it means they satisfy no multiplicative relations, which in essence is equivalent to the uniqueness of prime factorizations. And of course multiplication is the operation used to define primes in the first place, so any other kind of natural operation to be considered must somehow derive from multiplication one way or another.
I don't see any such group operation; do you? The operation $p*q:=\lfloor\frac{pq}{2}\rfloor$ seems pretty meaningless to me. What significance is it supposed to have? Or is it just randomly picked?
As mentioned in the comment, you can turn any countable set $X$ into a group isomorphic to an equal-size group $G$ by simply invoking a set-theoretic bijection $X\leftrightarrow G$. Then you simply need to relabel the elements of $G$'s multiplication table with the corresponding elements of $X$ to get the multiplication table of $X$ with its newfound group structure. This kind of construction (called transport of structure) is completely blind to any features that $X$ may have beyond just being a barren set, so there's no way we could consider this a "group structure on $X$" if the object $X$ already has special established meaning.
There is another type of structure we could identify primes as - a topological space, or more algebraically, an affine scheme. Algebraic geometry is the right context to understand what this perspective is all about, but it's quite abstract and sophisticated.
*The fundamental theorem of arithmetic (existence and uniqueness of prime factorizations for integers) is taken for granted too often as an obvious fact. It isn't as obvious as you might think; there are number rings which do not have prime factorizations for all elements and it's hard to point to any obvious feature on the surface that tells us when/why they are factorial or not.
Best Answer
The collection of positive real numbers (and even real numbers without zero) is a group. However, once you append zero, the resulting set is no longer a group for exactly the reason your are suggesting.
One interesting thing about the positive real numbers, $(\mathbb{R}_+,\cdot)$, is that they are isomorphic to the reals with addition, $(\mathbb{R},+)$. This can be seen through the logarithm, $$\log(a\cdot b) = \log(a) + \log(b).$$ Note also that $\log(1)=0$, that is the logarithm identifies the identity elements between the two different groups.