Finding an example of a set $G$ which is not a group Suppose $G$ is a set and $\cdot$ is a binary operation on $G$ such that there exists an $e\in G$ such that $a\cdot e=a$ for a in $G$ and given $a\in G$, there is a $y(a)\in G$ such that $y(a)\cdot a=e$.
I wish to find a $G$ such that $G$ is not a group under $\cdot$. 
An obvious place to start looking would be sets with some non-commutative operation. An example that comes to my mind would be that of square matrices under multiplication but it seems that if a square matrix is left invertible it has a right inverse as well, so this example doesn't work.
 A: A group doesn't need to be commutative, but we do require associativity. One example would be the integers under subtraction.
You can take $e=0$, and $y(a)=a$, so $\langle \mathbb{Z}, - \rangle$ satisfies your requirements, but is not a group.
A: As an addendum, note that your question asks for an example of a binary operation which has an identity (two-sided) and left inverses for every element. The two axioms which are missing here are the existence of two-sided inverses and associativity. The following fact shows that associativity must fail in such an example:
Let $(G, \cdot)$ be a set and binary operation, such that $\cdot$ is associative, and $G$ contains a two-sided identity for $\cdot$, and every $a \in G$ has a left inverse. Then $G$ is a group.
Proof: Let $a \in G$. Let $b$ be a left inverse for $a$. Let $c$ be a left inverse for $b$. Then $c(ba) = c(e) = c$, but $(cb)a = (e)a = a$. By associativity, we therefore have $c=a$, which makes $b$ a two-sided inverse for $a$.
A: For an example that is commutative (but not associative), let $G$ be any set with more than $2$ elements and define $a\cdot e=e\cdot a=a$ for all $a\in G$ as usual and  
$$a\cdot b=e\quad\text{if }a,b\not=e$$
Note that inverses here are not unique.
Another non-associative (but in this case non-commutative as well) example:  Let $G$ be the positive integers, let $y(a)=e=1$ for all $a\in G$, and let $a\cdot b=a^b$.
