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Here is the problem I am attempting to solve: Let $S$ be a set with a composition law $\cdot$ possessing the following properties:

$$(i)~x\cdot y = y\cdot x~\forall~x,y\in S$$ $$(ii)~x\cdot (x\cdot y) = y~\forall~x,y\in S.$$

Now define a new composition law by fixing $e\in S$ to be $$x+y = e\cdot(x\cdot y).$$

Prove that $+$ is commutative, that $e$ is the unique identity, that $a+x=y$ has a unique solution for all $x,y\in S$. Then find a condition on $\cdot$ so that $+$ is associative if and only if your condition holds. Prove your claim.

I have completed everything except that there is a unique solution to $a+x=y$. I have one solution: If I choose $a= x\cdot(y\cdot e)$, then $a+x = y$, as is verifiable. My question is this, how do I show that this is unique? If we were working over a group, I could use associativity easily. However, I am effectively told this is not an associative structure. I did discover (and prove!) that $+$ will be associative, hence $(S,+)$ will be a group, iff $z\cdot (e\cdot(x\cdot y)) = x\cdot (e\cdot (y\cdot z)).$

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Hint: If $a+x=y$, then $$y=e\cdot(a\cdot x)=e\cdot(x\cdot a).$$ What operations can you perform to this equation to end up with $a$ alone on the right-hand side?

A full answer is hidden below.

Multiplying by $e$ and then by $x$ on the left, you get $$x\cdot(e\cdot y)=x\cdot(e\cdot(e\cdot(x\cdot a)))=x\cdot(x\cdot a)=a.$$ So the only possible solution to $a+x=y$ is $a=x\cdot(e\cdot y)$.

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    I do not understand how you have proved this is the only solution. From $y$ you have generated a solution. For example, if $a+x =y$, then $x\cdot (a+x) = a$, but why is this not really a different solution, aside from the fact that it is circular? Could this "definition" of $a$ be used to prove that $a=x\cdot(e\cdot y)$? – 2017-01-11
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    This proves that if $a+x=y$, then $a=x\cdot(e\cdot y)$. So any value of $a$ that satisfies $a+x=y$ must be equal to $x\cdot(e\cdot y)$. You might define $a$ in some different way, but no matter how you define $a$, if $a+x=y$, then $a$ is equal to $x\cdot (e\cdot y)$. (This doesn't prove that this value of $a$ actually does satisfy $a+x=y$, but you've already done that part.) – 2017-01-11
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    I understand now. In fact, this was how I determined my value of $a$. Thank you for your assistance. – 2017-01-12