Find a binary operation $m$ on $S = \mathbb Z$ which has an identity, is associative and satisfies: $$∀x, y, z ∈ S, m (x, y) = m (x, z) ⇒ y = z$$ but which does not form part of a group structure on $S$.
I have tried to do it this way but I'm unsure if it makes sense specially with the inverse,
Identity: We want to find an element $e$ in $\mathbb Z$ such that for any $x \in \mathbb Z$, $m(x, e) = x$ and $m(e, x) = x$. Let's solve for $e$ in $$m(x, e) = x: x + e + 1 = x\Rightarrow e + 1 = 0 \Rightarrow e = -1$$ So, the identity element is $-1$.
Associativity: $$m(m(x, y), z) = m(x + y + 1, z) = x + y + z + 2\\ m(x, m(y, z)) = m(x, y + z + 1) = x + y + z + 2$$ Associativity holds. Satisfying the given condition: $m(x, y) = m(x, z)$ implies that $x + y + 1 = x + z + 1$, which simplifies to $y = z$.
Now, let's check whether this operation forms a group structure on $\mathbb Z$:
Closure: Addition is a closed operation on integers, so $m(x, y) = x + y + 1$ is also closed. Identity: We found that the identity element is $-1$. Inverses: For each element x, we need to find an element $y$ such that $m(x, y) = -1$ and $m(y, x) = -1$: $$m(x, y) = -1 \Rightarrow x + y + 1 = -1 \Rightarrow y = -x - 2\\ m(y, x) = -1 \Rightarrow y + x + 1 = -1 \Rightarrow y = -x - 2$$
You example is incorrect: the integers equipped with $x+y+1\:is$ a group. The structure you are looking for is a left-cancellative monoid. The set of nonnegative integers under addition is a cancellative monoid since positive integers do not have an additive inverse. Next, we need to find an isomorphism between $(\mathbb{N},+)$ and $(\mathbb{Z},\otimes)$, where $\otimes$ is a new binary operation compatible with the isomorphism. For example, let the isomorphism $f:\mathbb{Z}\to\mathbb{N}$ be defined as $$f(n) = \begin{cases} -2n & n \leq 0 \\ 2n-1 & n > 0 \end{cases}$$
The identity in $(\mathbb{Z},\otimes)$ would be $0$, and you could finish by working out the explicit form of the operation $\otimes$.