It is given that $\frac{ds}{dt} = \frac{d\theta}{dt}$, i.e. the time derivative of s and $\theta$ are equal to each other.
Does it follow that for small values of $t$, $\Delta s ≈ \Delta \theta$? My thought process for this was that if I multiply $dt$ on both sides, I am left with just the value of $ds=d\theta$, which I can rewrite as $\Delta s ≈ \Delta \theta$. Would I be correct in thinking this?
The answer depends on a necessary clarification: does the equation
$$\frac{ds}{dt} = \frac{d \theta}{dt}$$
hold for all $t$, or just for a particular value of $t = t_{0}$?
If the equations holds for all $t$ then the equation can be integrated and we will have
$$ s(t) = \theta(t) + C $$
where $C$ is some constant. I don't think this is what you are asking about?
If the equation only holds for a given instant $t=t_0$ then it is indeed valid to say
$$ \Delta s \approx \Delta \theta $$ for a small time interval
$$ t \in ( t_0, t_0 + \Delta t ) $$ where $\Delta t$ is small. (Note that in your post you say "for small values of $t$" where I guess you mean small values of $\Delta t$.)
(some) Mathematicians would probably scoff at "multiplying by $dt$ on both sides" but I think many physicists do this in practice. We just need to know what we actually mean when we do it.
