Is this structure a group? I'm currently working on a problem where a nonempty set $S$ has $\cdot$ an associative and cancellable operation. It is a bit ambiguous but when it says operation I feel the problem is talking about a binary operation, then this set would be a cancellable semigroup. 
$S$ also has the property that for every element the set of its powers is finite. Then, every element in $S$ has finite order, since the order is definided as the cardinality of such set.
The question is, 

Should $S$ be group?

When reading up on this question, I found out that if $S$ is itself finite, then $(S,\cdot)$ is a group. I tried setting up a similar function $s\mapsto as$, but I couldn't find a right inverse given that the elements had finite order.
The thing is that, if every element has finite order, must there be an identity element? I feel like there would be a lot of identities, $s^0$. But I haven't found a way to prove uniqueness or that it even exists.
I also thought of a structure where all elements have finite order but is infinite and came about with $\Bbb Q/\Bbb Z$. Now, this set is that it is an infinite group where all elements have a finite order. The problem is, it's already a group.
I can't seem to find a counterexample or give a proof to the problem. Any hints or solutions are kindly appreciated.
 A: Choose some element $s\in S$.  Since $s$ has finite order, there exist positive integers $a<b$ such that $s^a=s^b$; let $n=b-a$.  Now consider any $t\in S$, and observe that $s^as^nt=s^bt=s^at$.  Cancelling $s^a$, we see that $s^nt=t$.  Similarly, $ts^n=t$.  Thus $s^n$ is an identity.  It follows that $S$ is a group (indeed, the argument above shows that any element of $S$ has some power which is the identity).
More generally, a similar argument shows that if a semigroup $S$ has elements $x$ and $y$ such that $xy=x$ and $x$ is cancellable, then there is an identity element (namely $y$).  (To show $y$ is a right identity, observe first that $xyx=x^2$ and so $yx=x$.)
A: so the set of $A= \{a^k|k\in \mathbb N\}$ is finite.  It's easy to show that $A=\{a,a^2, .... a^m\}$ some $m$ and then $a^{i}; 1\le i\le m$ are distinct.
Now $a^{m+1} = a^k$ for some $1\le k\le m$.    If $k > 1$ then $a^m = a^{k-1}$ which is a contradiction.  So $a^{m+1} = a$ and $a^ia^m = a^ma^i= a^{m+1}a^{i-1} =aa^{i-1} = a^i= a^i$ so $a^m$ acts as an identity element on all $a^i$.
Similarly for $b\in S$ there is a $b^n$ that acts as an identity on $b^i$.  $a^{2m}b= a^mb^{n+1}$.  so $a^mb = b^{n+1} = b$ and likewise $ba^m = b^{n+1}a^{2m}$ so $b = b^{n+1}a^m = ba^m$. So $a^m$ will also act as an identity element on $b$.
