I was thinking about the following:
Does there exist real numbers $a,b, c, e, f$ and integers $m,n$ (where $m \geq -1$ and $n \geq -1$ with $m+n \geq 1$) such that $(-a + bc)^m (\frac{1}{a} + ef)^n = 1$. Since this is a perturbation to $(-a)^m (\frac{1}{a})^n=1$, we have that $bc,ef$ are small non-zero quantities.
We also have the information that $|a|<1$ and that there exist no integers such that $(-1)^m a^{(m-n)}=1$. The quantities $b,c,e,f$ are small.
I am trying to think about how to approach this question? Maybe a binomial expansion and neglecting some small quantities help (but it seems like it can get complicated)?
$\textbf{EDIT}$
$m,n$ are distinct integers. The case $(-a + bc)^m (\frac{1}{a} + ef)^n = 1$ can be thought of as a small perturbation of the case $(-a)^m (\frac{1}{a})^n = 1$. We further have the information that the perturbed modulus argument that is $|-a + bc|<1$ and $|\frac{1}{a} + ef|>1$.
$a = \frac 12$, $b=\frac 32$, $c=1$, $e =-1$, $f=1$, $m=n=1$ works. (Many more examples exist.)