$c_n=a^n-b^n$, $S=\{c_1,c_2,...\}$ is finite then $0 \in S$ Statement:

Let $  (R,+,\cdot)  $ be a ring that has at least two invertible
elements. Let $a,b$ be two invertible elements such that
$$ord(a)=ord(b)=+\infty$$ Let consider the following sequence
$$c_n=a^n-b^n$$  Let $S=\{c_1,c_2,...,c_m,...\}$ a set of elements
from $R$.
Prove that if $S=\{c_1,c_2,...\}$ is finite then $$0 \in S$$

Attempt:
I obtained that:
$$a^n-b^n=c_n$$
then $$a^{n+1}-ab^n=ac_n$$
and $$a^{n+1}-b^{n+1}=c_{n+1}$$
then
$$(b-a)b^{n}=ac_n-c_{n+1}$$
Since $S$ is finite then $$M=\{ac_n-c_{n+1},n \in \mathbb Z_+\}$$ is finite.
So $$\{(b-a)b^n,n \in \mathbb Z_+\}$$ is finite.
Then there exists $i$ and $j$ such that $$(b-a)b^i=(b-a)b^j$$
How should I continue with this. I would highly appreciate somebody's help.
 A: Modified to NOT assume $R$ is commutative
Continuing from where you left off, you could use your line of reasoning to show the following: For any positive integer $l$ [not just $l=1$], there is a positive integer $k_l$ such that
$$(a^{k_l}-1)c_l = 0.$$
[Indeed for each such $l$ there exists distinct integers $i$ and $j$ such that $a^ic_l = a^jc_l$, and assume $j >i$. Then $k_l=j-i$ works.] Furthermore, $(a^k-1)c_l =0$ for any $k$ that is a multiple of $k_l$, because even in a noncommutative ring, $(a^{C_1}-1)$ and $(a^{C_2}-1)$ commute for any two integers $C_1$,$C_2$. So let $T$ be a finite set of integers $l$ such that for all $m \in \mathbb{N}$ there is an $l \in T$ such that $c_m=c_l$, and set
$$k = \prod_{l \in T} k_l.$$
Then
$$(a^k-1)c_m = 0 \ \forall m \in \mathbb{N}.$$
In fact, for any multiple $K$ of $k$, it follows that
$$(a^K-1)c_m =0 \ \forall m \in \mathbb{N}.$$ Thus we conclude with the following:
Claim 1 There is a positive integer $k \in \mathbb{N}$ such that the equation $$(a^K-1)c_m=0 \ \forall m \in \mathbb{N}$$ holds for all multiples $K$ of $k$.
We also make the following claim:
Claim 2 There exists a multiple $K$ of $k$ and a positive integer $n$ such that the equation $c_{n+K}=c_n$ holds.
Proof: Let $N$ be the number of distinct $c_l$; $l \in \mathbb{N}$. Then for any positive integer $m$, at least two of $c_m, c_{m+jk}$; $j=1,2,\ldots, N$; must be the same by the Pigeonhole Principle. ■
So, using Claims 1 and 2, let $K$ be a multiple of $k$, where $k$ is as in Claim 1, and let $n$ be an integer such that the equation $c_n=c_{n+K}$ holds.
Then simple algebra gives
$$a^Kc_n +a^Kb^n -b^{n+K} = c_{n+K}.$$
plugging into this the equation $c_n=c_{n+K}$ and rearranging gives
$$(a^K-1)c_{n+K} + c_Kb^n = 0.$$
But, as $K$ is a multiple of $k$ as in Claim 1, the equation $(a^K-1)c_{n+K}=0$ holds, so we are left with the equation
$$c_Kb^n = 0.$$
As $b^n$ is invertible, it follows that the equation $$c_K=0$$ must hold. ■
