First of all, when you say that the sequence follows the rule that you specify just by looking at its first entries, you are making a guess ... a reasonable guess, of course, but a guess nevertheless.
Anyway, let's just say that the rule you indicate does indeed specify the sequence we are interested in. Is it periodic?
Your observation that given that the number of pairs of numbers is finite, and hence that there has to be a repeat of some a pair of numbers at some point is correct.
However, there is no need to worry about the predecessors: once you have a repeat of $x,y$ in your sequence, then, given the rule, from then on everything will be the same again, and that is all you need to show it is cyclic.
In fact, there are many sequences that are cyclic, but where the cycle is preceded by a little 'start-up' that does not get repeated ... i.e. where its graph looks like a '6', not a '0'. For such sequences, you would not be able to prove that the predecessors are always the same as well, because that would not be true. But, you still have that cycle, and that makes it periodic.
In sum, the sequence will be periodic, though it may not go back to $(1,1)$
That said, it is actually easy to show that the sequence will go back to $(1,1)$, for all the predecessors will be the same: if you have
$$b,x,y$$
and
$$f,x,y$$
then it must be the case that
$$b+ x =y \mod 10$$
and
$$f+x=y \mod 10$$
and from that it immediately follows that $b=x$ given that $0 \le b,f \le 9$
Once you have established that $b=f$ it likewise follows that $a=e$ ... but you really don't need to show that anymore: once you have shown the immediate predecessor of two numbers is fixed, then all predecessors are fixed. In short, from any $a_n$ and $a_{n+1}$, both $a_{n+2}$ and $a_{n-1}$ are fully determined.
