Are associativity and solvability for $x$ in all $x+a=b$) sufficient laws for a group? Typically a group is defined as a set with a single binary operation mapping pairs of elements to elements of the set.  The operation is associative, and there exists a neutral element $0$ such that $0+a=a$.  
It occurs to me that if we have a set with an associative binary operation defined in it, and declare that for every equation of the form $x+a=b$ there exists an element $x$ satisfying the equation, we can show the existence of both a neutral element, as well as an inverse to every element.
That is, the solution to $x+a=a$ will be $0$.  Once we demonstrate the existence of $0$, we require solutions to all equations $x+a=0$.  
Are these two assumption sufficient to characterize a group?
 A: Unfortunately, it isn't, even if we add the assumption that the underlying set is non-empty. The simplest counterexample I can think of is to take any set $S$ with more than one element and define an operation $*$ on $S$ by $a*b=a$ for all $a,b\in S.$ Then $$a*(b*c)=a=a*c=(a*b)*c,$$ so $*$ is associative. Moreover, for any $a,b\in S,$ we have $x=b$ as a solution to $x*a=b,$ as you desired--in fact, $x=b$ is the unique solution! However, this operation cannot have an identity. If there were such, say $e,$ then taking $a$ and $b$ to be distinct elements of $S,$ we obtain the absurdity $$a=e*a=e=e*b=b.$$
On the other hand, let's suppose we imposed the following conditions:


*

*$S$ is a non-empty set.

*$\star$ is an associative operation on $S.$

*For any $a,b\in S,$ there exists some $x\in S$ such that $x\star a=b.$

*For any $a,b\in S,$ there exists some $y\in S$ such that $a\star y=b.$
Then $S$ is a group under the operation $\star.$
To prove it, we begin by proving that the identity exists. That is, we must show that there is some $e\in S$ such that for all $a\in S,$ $a\star e=a=e\star a.$
Take any $a\in S.$ We know that there exists $x\in S$ such that $x\star a=a.$ We'd like to show that $x\star b=b$ for all $b\in S.$ Take any $b\in S.$ We know there is some $y\in S$ such that $a\star y=b.$ Then $$x\star b=x\star(a\star y)=(x\star a)\star y=a\star y=b,$$ as desired. Thus, we've shown that there is at least one left-identity element in $S.$ A similar proof shows that there is a right-identity element in $S.$
Suppose that $l,r\in S$ are left- and right-identity elements. That is, for all $x\in S,$ we have $l\star x=x=x\star r.$ Then with $x=r$ we see that $l\star r=r,$ and with $x=l,$ we see that $l\star r=l.$ Thus, every left-identity element is a right-identity element, and vice versa, so we've proved that there is at least one identity element. We can (and should) further prove that there is exactly one identity element, but I leave that to you.
Finally, we prove the existence of inverses. Letting $s\in S$ be arbitrary and letting $e$ be the identity element, we know that there exists $x\in S$ such that $x\star s=e.$ It remains only to show that $s\star x=e.$ We know that there is some $y\in S$ such that $(s\star x)\star y=e.$ Since $e$ is the identity element, then $$s=s\star e=s\star(x\star s)=(s\star x)\star s,$$ so we see that $$s\star x=\bigl((s\star x)\star s\bigr)\star x= (s\star x)\star (s\star x),$$ and so
\begin{eqnarray}e &=& (s\star x)\star y\\ &=& \bigl((s\star x)\star(s\star x)\bigr)\star y)\\ &=& (s\star x)\star\bigl((s\star x)\bigr)\star y)\\ &=& (s\star x)\star e\\ &=& s\star x,\end{eqnarray} as desired.
