Q:

Properties of x*y = |x+y| in set R

This is the second exercise in the second set of chapter 2 in Pinter's A Book of Abstract Algebra.
It is asked to check for commutativity, associativity, identity element in $\mathbb{R}$ and inverse elemente in $\mathbb{R}$ for each described operation, in this case i have a question about $x*y=|x+y|$.
My first question lies in the identity check, this is what i did: 
$ x*e = |x + e| = \sqrt {(x+e)^2} = x$
so $e=0$, checking for $e*x $ gives same thing.
Alright, but if x is negative then the operation will not give back a negative x, but a positive x, thus this operation is not injective. The book explicity says that for something to be an operation it needs to be uniquely defined for every  $x\, \in\, \mathbb{G}$
I thought about using $ \pm $ but it doesn't make sense to me in that case.
Also, i don't have access to the book's solution. 
My second question is about the associativity, i did:
$ (x*y)*z = |x+y|*z = ||x+y|+z| = \sqrt{(\sqrt{(x+y)^2} + z)^2} = \sqrt{(x+y)^2 + 2z\sqrt{(x+y)^2} + z^2}$ 
$ x*(y*z) = x*|y+z| = |x + |y+z|| = \sqrt{(\sqrt{(y+z)^2} + x)^2} = \sqrt{(y+z)^2 + 2x\sqrt{(y+z)^2} + x^2}$ 
Using computational resources i get that $(x*y)*z = x*(y*z)$ will be true under certain conditions, but i just can't visualize it. 
Thanks.

A:

How can $0$ be an identity?
$-1\ast 0 = |-1 + 0| = |-1| = 1 \neq -1$.
It is not hard to see that there is NO real number $e$ that will yield:
$x\ast e = x$ for any $x < 0$.
Also:
$[(-1)\ast(-1)]\ast 1 = |(-1) + (-1)| \ast 1 = |-2|\ast 1 = 2\ast 1 = |2+1| = |3| = 3$, but:
$-1 \ast[(-1)\ast 1] = -1 \ast |-1 + 1| = -1 \ast |0| = -1 \ast 0 = |-1 + 0| = |-1| = 1$.
All is takes is ONE counter-example for disproof.
Without an identity, how can you have inverses?

