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.
You're on the right track, but are indeed missing a few small things.
First, note that you can only get that highest possible sum of $69$ if $A =\{ 9,10,11,12,13,14 \}$. But any non-empty subset of that set has a sum of at least $9$, meaning that for any non-empty subset of this particular set, the sum of its members is at least $9$, and at most $69$, meaning that there are only $61$ possible sums for this subset, rather than $69$. Hence, the pigeonhole principle can be applied to this particular set.
OK, but how can we be sure the pigeonhole principle can be applied to any set $A$?
Well, note that the key observation we made regarding the specific set above was that the difference between the highest sum and lowest sum of non-empty subsets was $60$, and that was because that set contained the numbers $10$ through $14$. And now note that for any set, this difference between the highest and lowest set can never be greater than that, and is for most sets smaller than that.
So, we can say that for any set $A$, there are at most $61$ possible sums for the non-empty subsets of $A$, but there are $63$ non-empty subsets of $A$. So yes, the pigeonhole principle can be applied to any set $A$ to show that there must always be two subsets with the same sum.
Best Answer
Suppose $S$ contains two equal elements. Then the subsets of $S$ comprising only these elements have equal sums, so the sums are not all distinct.
Now suppose $S$ does not contain two equal elements, i.e. the elements of $S$ are all distinct.
Let $x$ be the smallest element of $S$.
Then the smallest possible value of $S_A$ is $x$.
And the largest possible value of $S_A$ is $x+10+11+12+13+14=x+60$.
Since $x<S_A<x+60$, there are $61$ possible values of $S_A$.
And since there are $2^6-1=63$ non-empty subsets, these cannot all have distinct sums since there are only $61$ possible values available for the sum.