Handling the real numbers, multiplication, and zero as a group A group can be formed using a set and a binary operator on elements of a set. Consider {$\mathbb{R}$, x}, the real numbers $\mathbb{R}$ and the multiplication operator x. There are nine product rules. Often letters are used for product rules, so think of p->+1, n->-1, and z-> 0:


*

*R1: p x p = p

*R2: n x n = p

*R3: p x n = n

*R4: n x p = n

*R5: z x z = z

*R6: z x p = z

*R7: z x n = z

*R8: p x z = z

*R9: n x z = z


The first four rules are the product rules for the cyclic group $Z_2$, or the sign group. By themselves, R1 and R5 are different representations of the trivial group $Z_1$. These 4 qualities are required to show that {$\mathbb{R}$, x} is a group:


*

*G1: Closure: $a$, $b$ in $\mathbb{R}$ and $a x b$ is in $\mathbb{R}$

*G2: An identity $I$ exist for all in $\mathbb{R}$: $a I = I a = a$

*G3: An inverse exists for all in $\mathbb{R}$: $a a^{-1} = I$

*G4: The operation is associative: $(a x b) x c = a x (b x c)$
I will presume G1 and G4 (minimal requirements for a semigroup) and focus on G2 and G3.
For G2, the multiplicative identity is $1$, so this works for all elements in R, even zero since $0x1=1x0=0$.
It is well known that G3 does not work, that there is no $a^{-1}$ such that $0 x a^{-1} = 1$. To deal with this, it is common to consider the set $\mathbb{R}$\ $0$, the real numbers without zero as a group: {$\mathbb{R}$\ $0$, x}. This is in fact a group $Z_2$, using only rules R1-R4.
What is troubling is that $\mathbb{R}$ \ $0$ cannot have the rules R5-R9. With no z, the rules are undefined. Zero is not the same as undefined. Perhaps there have been mathematicians that have been upset by this observation that the set $\mathbb{R}$ \ $0$ makes product rules R5-R9 suspect.
Here is a possible resolution. Here is an an alternative to G2 where differences are denoted with stars:
G2: An identity I(a) exist for every in R and Z1: a I(a) = I(a) a = a 
The addition of and Z1 could be done to the original G2 without changing its meaning since $1x1=1$ by rule R1. The trivial group $Z_1$ and the addition operator has only one representation: $0+0=0$. The trivial group $Z_1$ and the multiplication operator has two representations: $1x1=1$ and $0x0=0$. The switch from all->every allows one to use a different representation of the trivial group to be the identity. This is why I->I(a): the representation of the identity depends on the value of a. If and only if $a=z$, then the identity equals z, otherwise it is the far more familiar 1. The multiplicative inverse of z is z because $axa^{-1}=I(a)$ becomes: $0x0=0$. All 9 product rules will work for every real number, zero included.
Thus {$\mathbb{R}$ \ $0$, x} is a group, {$\mathbb{R}$, x} is not a group, but {$\mathbb{R}$, x} is a group based on the modifications to G2 discussed above. Are there mathematicians who have argued along these lines?
 A: I've never heard of a mathematician who disliked the fact $(\mathbb R,\cdot)$ is not a group.  As far as I know, we are almost all content that it is a commutative monoid consisting of an abelian group with a zero element adjoined.
I know people think rings with local identities (that is, for each element, there is an element that acts like an identity for it), and I assume that people study semigroups with local identities too, which seems to be what you're angling at.
By the way, replacing "for all" with "for every" as you have does not really change the meaning that way (if that is what you intend.)  Rather, you have to change the order of the quantifiers:  "For each $a$ there exists an $I(a)$ such that..." .
A: It is a bad idea to reuse the same name for a different algebraic structure. A group is a useful algebraic structure, and its definition is fine as it is.
However the algebraic structure you are talking about has been considered by some mathematicians, it just (for good reason) is not called a group. Instead it is called a completely regular semigroup.
A: I agree you should not refer to this as a group. A group already has a well understood meaning, and trying to change the definition of a well accepted term will just cause confusion.
Depending on precisely what you mean (it is unclear what your modified G2 is stating), you may be noticing that $(\mathbb{R},\cdot)$ is an inverse semigroup. (EDIT: Or more strongly, a Clifford semigroup which is a completely regular inverse semigroup.)
An inverse semigroup $(A,\cdot)$ is a semigroup where each element has a unique "inverse" in the following sense:
$$\text{For all } x \in A, \text{there exists a unique } y \in A \text{ such that } x = x\cdot y\cdot x \text{ and } y = y\cdot x\cdot y.$$
So when considering $(\mathbb{R},\cdot)$ as a semigroup, for non-zero real values, the semigroup inverse is the usual value, and the semigrouop inverse of 0 is 0.
Please take note that these are not normal inverses. For example if you had the equation
$ab=c$, it is not always possible to solve for $b$ using $b=a^{-1} c$, where $a^{-1}$ is the semigroup inverse.
In this case there is only one oddball element, so to make this distinction even clearer, consider $\mathbb{R}^2$ and define the operation $(a,b)\cdot(c,d) \rightarrow (ac,bd)$. This is an inverse semigroup.  However there are now non-zero elements $a,b,c$ where $ab=c$, yet $b \ne a^{-1} c$. For example, $a=(2,0),b=(2,3),c=(4,0)$. Therefore these "inverses" cannot be used like group inverses, even in this simple linear equation.
So these "inverses" do not give you a nice division operation in the way you may have hoped. Just like you still cannot divide 9 by 0 in your $(\mathbb{R},\cdot)$ inverse semigroup.
You note:

...this observation that the set R \ 0 makes product rules R5-R9 suspect.

No.
Nothing you said made the multiplication by 0 in the reals "suspect" just because the multiplicative group does not include the additive identity.
If you want to consider a structure including 0, it would be more appropriate to view the reals as a division algebra. This includes the interaction between addition and multiplication, in particular the distributive law. Now $(\mathbb{R},+)$ is a group involving all the real numbers, with the additive identity being 0. The distributive law tells us multiplication and addition interact such that the additive identity cannot be part of the multiplicative group. There is nothing mysterious about this, and it doesn't mean we are unable to discuss 0 with multiplication. The distributive law, and the fact that zero is the additive identity tells us that for any $a$:
$$ 0\cdot a = (0+0)\cdot a = 0\cdot a + 0\cdot a $$
Which is only possible if $0\cdot a$ is the additive identity and thus 0 times any number must be 0.
If you want to more meaningfully include how 0 (additive identity) interacts with the multiplicative group, you should focus on the structure that meaningfully combines them: the real numbers as a division algebra.
