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I have a polynomial equation in $x$ and $y$,

$$
(a-b)(xy+1) + (ab+1)(y-x) = 0.
$$
What transformation can act on $x$ and $y$ so that any point that satisfies this equation is mapped to another point that satisfies it? If there is a general way to find these transformations given a polynomial equation, that would also be nice to know.

the substitutions
$$ x = u + fracab+1b-a ; , ; $$
$$ y = v - fracab+1b-a ; , ; $$
take us to the hyperbola
$$ uv = C $$
where $C = C(a,b)$ is a constant.

One motion, for real $t neq 0,$ is
$$ (u,v) mapsto left(tu, fracvt right) $$

You have $a,b,x,yinmathbbR$ ($mathbbR$ can be replaced by $mathbbC$ in every occurrence in this answer) such that
$$fraca-bab+1=fracx-yxy+1.tag*$$
Let $c:=arctan(a)$, $d:=arctan(b)$, $u:=arctan(x)$, and $v:=arctan(y)$. Consider $c$, $d$, $u$, and $v$ as elements of $mathbbR/pimathbbZ$. Then, we have
$$tan(c-d)=tan(u-v),.$$
Consequently, $c-d=u-v$.

Now, any transformation $T:mathbbR/pimathbbZtomathbbR/pimathbbZ$ of the form $$T(t)=t+qtext for all tinmathbbR/pimathbbZ,,$$ for some fixed constant $qinmathbbR/pimathbbZ$ satisfies
$$T(t_1)-T(t_2)=t_1-t_2,$$
for all $t_1,t_2inmathbbR/pimathbbZ$. Consider the function $f:mathbbRtomathbbR$ sending $$smapsto tanbig(arctan(s)+qbig)=fracs+p1-sptext for all sinmathbbR,,$$
where $qinmathbbR$ is a fixed constant and $p:=tan(q)$. Then, if $a,b,x,yinmathbbR$ satisfy (*), then
$$fraca-bab+1=fracf(x)-f(y)f(x),f(y)+1$$ too.

In fact, only functions $f$ of the given form satisfy the requirement. Thus, all solutions $(x,y)inmathbbRtimesmathbbR$ to (*) is of the form
$$(x,y)=left(fraca+p1-ap,fracb+p1-bpright)$$
for some $pinmathbbR$ such that $pneqdfrac1a$ and $pneq dfrac1b$.