This looks pretty good, assuming your ring is commutative (as noted in a comment above). Usually we discuss closure first, because without it, there's not really anything to talk about, but that's not essential.
What's required to establish is just: 0) The product of two units is a unit - you discovered that proof, and it will be fully general if you use $b'a'$ as your inverse instead of $a'b'$; 1) Multiplication is associative - free from the ring axioms, as you noted; 2) The multiplicative identity from the ring is a unit and serves as our group identity - got that; 3) The multiplicative inverse of a unit exists and is another unit - free from the definition of "unit", as you noted.
You established all of those things, so you're good. I'm sure the wording could be tightened up, but that will come with time. The only thing I'd really take issue with now is that the product argument is a little clunky. There's no need to assume that $ab\neq e$. You can really just say:
"Let $a$ and $b$ both be units, with multiplicative inverses $a'$ and $b'$. Then the element (b'a') is a multiplicative inverse for $ab$, so $ab$ is also a unit."
The statement of the associativity condition is wrong; it should be
$$(a \ast b) \ast c = a \ast (b \ast c) .$$
Expanding the l.h.s. gives
\begin{align}(a \ast b) \ast c &= (a + b - ab) \ast c \\ &= (a + b - ab) + c - (a + b - ab) c \\ &= a + b + c - bc - ca - ab + abc .\end{align}
Now, do the same thing for the r.h.s. and verify that the expressions agree.
We cannot simply assume the existence of an identity and inverse. There are plenty of associative binary operations which lack one or both!
We can pick out the identity element using the definition: We must have, for example, that
$$a \ast e = a$$ for all $a$, and expanding gives
$$a + e - ae = a.$$
Can you find which $e$ makes this true for all $a$?
Likewise, to show that the operation admits inverses, it's enough to produce a formula for the inverse $a^{-1}$ of an arbitrary element $a$, that is, an element that satisfies $a \ast a^{-1} = e = a^{-1} \ast a$. As with the identity, use the definition of $\ast$ and solve for $a^{-1}$ in terms of $a$.
Remark If we glance at the triple product, which by dint of associativity we may as well write $a \ast b \ast c$, we can see the occurrence of the elementary symmetric polynomials in $a, b, c$, which motivates writing that product as $(a - 1) (b - 1) (c - 1) + 1$. Then, glancing back we can see that we can similarly write $a \ast b = (a - 1) (b - 1) + 1,$ which is just the conjugation of the usual multiplication (on $\Bbb R - \{0\}$) by the shift $s : x \mapsto x + 1$, that is, $a \ast b := s(s^{-1}(a) s^{-1}(b))$. Since multiplication defines a group structure on $\Bbb R - \{0\}$, that $\ast$ defines a group structure follows by unwinding definitions. For example, to show associativity, we have
$$(a \ast b) \ast c = s(s^{-1}(s(s^{-1}(a) s^{-1}(b))) s^{-1}(c)) = s(s^{-1}(a) s^{-1}(b) s^{-1}(c)) = s(s^{-1}(a)s^{-1}(s(s^{-1}(b) s^{-1}(c))) = a \ast (b \ast c) .$$
(Note that in writing the third expression without parentheses we have implicitly used the associativity of the usual multiplication.) You can just as well use this characterization to determine the group identity $e$ and the formula for the inverse of an element under $\ast$.
Indeed, this together with analogous arguments for the other group axioms shows that conjugating any group operation by a set bijection defines an operation on the other set.
Best Answer
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.