Is a semigroup $G$ with left identity and right inverses a group? Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group?
Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a left inverse. For $a \in G$ denote this left inverse by $a^{-1}$. Then
$$(aa^{-1})(aa^{-1}) = a(a^{-1}a)a^{-1} = aa^{-1}$$
and we can use the fact that 
$$cc = c \Longrightarrow c = 1$$ 
to get that inverses are in fact two-sided:
$$ aa^{-1} = 1$$
From which it follows that
$$a = 1 \cdot a = (aa^{-1})a = a (a^{-1}a) = a \cdot 1$$ 
as desired.
But in the scenario given we cannot use $cc = c \Longrightarrow c = 1$, and I can see no other way to prove this. At the same time, I cannot find a counter-example. Is there a simple resolution to this question? 

If $cc=c$ than $c$ is called idempotent element.
Semigroup with left unit and right inverse is called left right system or shortly $(l, r)$ system.
If you take all the idempotent elements of $(l,r)$ system they also form $(l,r)$ system called idempotent $(l,r)$ system. In such system the multiplication of two elements is equal to the second element because if you take $f$ and $g$ to be two such elements, and $e$ is the unit, than
$fg=feg=fff^{-1}g=ff^{-1}g=eg=g.$
So, each element in such a system is left unit and also right inverse and any such system is not a group cause no element can be right unit.
It is now also easy to see that, if you define the multiplication of two elements to be equal to the second one, you get idempotent $(l,r)$ system, which is obviously not a group.
For more details and some extra facts check the article from 1944. by Henry B. Mann called 
"On certain systems which are not group".
You can easily google it and find it online.
Reference to this article is mentioned somewhere at the beginning of the book 
"The theory of groups" by Marshall Hall.
