Yes you do need to show that y=x. Also, you have found an element that acts like an identity when multiplying a particular element a. But you need to find a single identity element for the whole of S.

So, for the given element a, you know that there is an element e such that ae=a, and there is an element f such that fa=a. Now let b be another element of S. There is an element y such that ya=b. Therefore be=yae=ya=b, and by a similar argument fb=b. This shows that e acts as a right identity for every element of S, and f acts as a left identity for every element of S. Therefore in particular e=fe=f. Hence e is an identity.

No, you can't do that, because you don't yet know that inverse elements exist. However, given an element a, you do know that there is an element x such that ax=e, and there is an element y such that ya=e. But then y=ye=yax=ex=x, so y=x is an inverse for a. So each element has an inverse, and now you just have to check that the group axioms hold.

Again, you don't know that inverses exist, so you can't use that argument.

Suppose that S contains n elements (n is finite). Given an element a in S, think about the set of products ax as x runs through S. There are n of these products and (by the cancellation property) they are all distinct. So they must constitute all the elements of S. Therefore, given an element b of S, one of those products ax must be equal to b. In other words, the equation ax=b has a solution. Similarly, the equation ya=b has a solution. So now we can use the result from problem 1 to conclude that S is a group.

Suppose S is the set of natural numbers, with ordinary multiplication as the semigroup operation. The cancellation property holds, but S is not a group.