To show $H$ is a subgroup you must show it's closed, contains the identity, and contains inverses. But if it's closed, non-empty, and contains inverses, then it's guaranteed to contain the identity, because it's guaranteed to contain something, say, $x$, then $x^{-1}$, then $xx^{-1}$, which is the identity.
$H$ is assumed closed, so if it contains $a$ and $b$, it contains $ab$. But $a$ and $b$ don't have to be different: if it contains $a$, it contains $a$ and $a$, so it contains $aa$, which is $a^2$. But then it contains $a$ and $a^2$ so it contains $aa^2$ which is $a^3$. Etc.
So it contains $a,a^2,a^3,a^4,\dots$. $H$ is finite, so these can't all be different, so some pair is equal, that is, $a^i=a^j$ for some $i\ne j$.
As for your last question, do you know any example of a group with a non-cyclic subgroup?
The simple answer to your question is that your choice of "pigeons" and "holes" didn't support a proof of the claim. That's not to say it didn't do something. As the commenters correctly pointed out, you showed that you can't have a set of $50$ numbers less than $99$ such that any two numbers in the $50$ sum to a unique number, relative to any other pair. If this is what the question had been asking, then your logic is absolutely sound.
But the quesiton wasn't asking that. It was asking for the existence of $99$ as one of the pairwise sums. And, in this case, the choice of pigeons and holes given in the book works, and yours doesn't.
The thing about the pigeon hole principle, or any other proof technique, is that it's a tool. And like any tool, it has to be applied the right way to solve the problem at hand. This is a great lesson for you as a mathematician- to see that there are many ways to apply a single tool to a single problem. It's not enough to say "The pigeon hole principle didn't work". You must say "The pigeon hole principle, with this choice of pigeons and holes, gives me this result, which doesn't seem to help." All the same, you mentally note it in case you need it later, and then you either change your choice of tool or try to apply the tool in a different way. Problem solving is an art, as much as a science.
Best Answer
In this example you are putting the subsets of $S$ (which all have sums) and placing them into the sums of $1$ to $69$. Think of the sums of $S$ as bins, we only have $63$ non empty subsets, and $69$ bins to put them in.