First of all, you must check that this process stays inside $\{1...99\}$. At each stage the iterated function $f$ adds a number between $0$ and $9$, so in order to have $f(x)>99$, we would need $x>90$. But for $x>90$, the units digit is always smaller, so you're always adding a negative value. Similarly, to have $f(x)<1$, we would need $x<10$, but then the units digit is always bigger, so you always add a positive value.
So, since we always stay within that finite set, we must get a repetition eventually. As soon as that happens, we'll enter a cycle. We have only to verify that the one you describe in your question is the only one. This can be checked explicitly by getting a computer to calculate $f(x)$ for $1\leq x\leq 100$:
Bigger image. Sorry about the messy graph, couldn't find good plotting software.
I somewhat doubt there's any more profound proof than that. It's a fairly arbitrary mapping as far as I can see. The fact that it will eventually repeat is something common to all mappings of a finite set to itself. Some of those mappings have $3$ cycles. Some have $18$. This one happened to have $1$. Sure, why not. Or, to put it another way, the non-trivial feature of your function is that it maps $\{1, ... 99\}$ into itself, and the cycle is just a corollary of that.
What I do find interesting is the symmetry of the above graph. It could be explained if we could prove that:
$$f(99 - x) = 99 - f(x)$$
That is, the involution $x\to99-x$ is an automorphism for this system (not sure if the word "automorphism" is actually used in this context, but if nothing else it's certainly an automorphism of the above directed graph). This is because this involution has the effect of replacing each digit in a two-digit number with $9$ minus itself, so $10x+y$ gets mapped to $10(9 - x) + (9 - y)$. So the difference between the two digits changes from $y-x$ to $(9-y)-(9-x)=x-y$, that is, it gets multiplied by $-1$. So if $x$ is a two digit number and $f(x)=x+d$, we have:
$$99-f(x)=99-(x+d)=(99-x)-d=f(99-x)$$