Does this almost-group uniquely define a group? Consider a quasigroup $(S,+)$ such that for every $a,b,c,d\in S$, $$(a+(b+c))+d=a+(b+(c+d)).$$
This is almost a group, but not quite. For instance, $(\mathbb Z,-)$ satisfies those axioms.
You can easily prove that for any element $x$, the operation $a+_xb:=a+(x+b)$ yields a group.
But I wonder: for any given $(S,+)$ and any pair $x,y \in S$, are $(S,+_x)$ and $(S,+_y)$ neccessarily isomorphic?
 A: A colleague of mine has been able to prove that this is, indeed, the case. His proof had some redundant steps, below is a polished version.
Theorem: If a quasigroup $(S,+)$ satisfies the axiom $(*)$:
$$ (a+(b+c))+d=a+(b+(c+d))$$
Then for any $x \in S$, operation $a+_xb:=a+(x+b)$ forms a group. Furthermore, any two such groups $(S,+_x)$,$(S,+_y)$ for given $(S,+)$ are isomorphic.
Lemma 1: $(S,+)$ has right-identity.
Proof: $(S,+)$ is a quasigroup, for any $c$ there exists $d$ such that $c+d=c$. Therefore:
$$(a+(b+c))+d=a+(b+c)$$
Let $x:=(a+(b+c))$. Since we've put no constraints on $a,b,c$, this can be any element of $S$. We have:
$$x+d=x$$
So $d$ is the right-identity. From this point on the right-identity will be denoted as $0$.
Corollary: From the existence of right-identity and the properties of a quasigroup, it trivially follows that any operation $+_x$ has a right-identity and right-invertability.
Lemma 2: $+_x$ is asociative.
Proof:
$(a+_xb)+_xc\overset{def}{=}(a+(x+b))+(x+c)\overset{(*)}{=} a+(x+(b+(x+c)))\overset{def}{=}a+_x(b+_xc)$
This establishes that $(S,+_x)$ indeed forms a group.
Lemma 3: $$0+(0+a)=a$$
Proof: $$x+a=(x+(0+0))+a\overset{(*)}{=}x+(0+(0+a))$$
Since $(S,+)$ is a quasigroup, the lemma follows.
Lemma 4: $$(x+y)+(0+z)=x+(y+z)$$
Proof:
$$(x+y)+(0+z)=(x+(y+0))+(0+z)\overset{(*)}{=}x+(y+(0+(0+z)))\overset{\text{L3}}{=}x+(y+z)$$
Now, we have sufficient tools to prove that $f(x)=k+x$ is a homomorphism from $(S,+_k)$ to $(S,+_0)$ for any $k \in S$
$$f(x)+_0f(y)\overset{def}{=}(k+x)+(0+(k+y))\overset{(*)}{=}((k+x)+(0+k))+y\overset{\text{L4}}{=}(k+(x+k))+y\overset{(*)}{=}\cdots$$
$$\cdots \overset{(*)}{=} k+(x+(k+y))\overset{def}{=}f(x+_ky)$$
This establishes homomorphism. Now, since $(S,+)$ is a quasigroup, $f(x)$ a bijection and therefore isomorphism. Since for any two $x,y \in S$, $(S,+_x)$ and $(S,+_y)$ are isomorphic with $(S,+_0)$ and isomorphism is obviously transitive, this proves the theorem.
Corollary: this theorem led to a surprizing rezult. All quasigroups $(S,+)$ satisfying the axiom $(*)$ can be represented as: $x+y=x \star f(y)$, where $(S,\star)$ is a group, and where $f(x)$ is an automorphism of said group such that $f(f(x))=x$. Furthermore, any operation $x \star f(y)$ forms a quasigroup satisfying the axiom $(*)$. The proof isn't hard, but it's beyond the scope of the original question, so I'll omit it.
