Too long to put in the comment box:
You got the set (multiplicative monoid actually) $X:= \mathbb Z [\frac16]^+$, and your function $f:X \rightarrow X$ on it, good.
It's also good you can compute the $n$-fold composition $f^n$ explicitly. It is true and neat that for all $x \in X$,
$$\lim_{n\to \infty} f^n(x) = x +2^{v_2(x)}3^{v_3(x)-1}$$
where the limit is taken w.r.t. the topology coming from $X \subset \mathbb R$.
On the other hand, it is true that $f(2x) =f(3x)=f(x)$ for all $x \in X$. This implies that $f$ naturally induces a function on the quotient $Q := X /\langle 2, 3 \rangle$, i.e.
$$\tilde f (x\cdot \langle 2,3 \rangle) := f(x) \cdot \langle 2, 3 \rangle$$
is well-defined and gives us a function $\tilde f :Q \rightarrow Q$, very good.
But to speak of convergence of any sequence in $Q$, one has to put a topology on $Q$.
The only natural candidate for that is the quotient topology. Alas, that topology is the indiscrete (sometimes called "trivial") topology, where the only open sets are the empty set and the full set $Q$.
In this topology, every sequence converges to any point ("at the same time").
So even though it is indeed true that
$$\lim_{n\to\infty}\tilde f^n(x\cdot \langle 2,3\rangle)= (x +2^{v_2(x)}3^{v_3(x)-1} )\cdot \langle 2, 3 \rangle,$$
this is true for the trivial reason that
$$\lim_{n\to\infty}\tilde f^n(x\cdot \langle 2,3\rangle)= a\cdot \langle 2, 3 \rangle$$
for all $a \in X$.
To see the issue in a clearer but similar example: Let $X = \mathbb R^+$. Take the quotient $Q:= \mathbb R^+/\mathbb Q^+$. Look at the sequence $$\sqrt2, \quad 1.4 \sqrt2, \quad 1.41\sqrt2, \quad 1.414\sqrt2, \quad 1.4142\sqrt2 \quad ...$$
You will agree that in $X$, this sequence converges to $2$. Now look at the corresponding sequence in the quotient $Q$. You will want to say, and it might be reasonable to assume, that it still converges to $2$ or rather, to its residue class $2\cdot \mathbb Q$ (which happens to be just $\mathbb Q$). But also, the sequence in the quotient is just
$$\sqrt2 \cdot \mathbb Q, \quad \sqrt2 \cdot \mathbb Q, \quad \sqrt2 \cdot \mathbb Q, \quad \sqrt2 \cdot \mathbb Q, \quad\sqrt2 \cdot \mathbb Q \quad ...$$
because all those factors $1, 1.4, 1.41,...$ were rational so get "swallowed" by $\mathbb Q$. So the sequence is actually constant, so it is also very reasonable to say it converges to the non-trivial residue $\sqrt 2 \cdot \mathbb Q$. So there you have it already converging to two different points in the quotient, and so far we have not even chosen a topology on $Q$, just demanded that it satisfies the very reasonable assumptions
a) If $\lim_{n\to \infty} a_n = a$ in $X$, then $\lim_{n\to \infty} \bar a_n = \bar a$ in $Q$.
b) For a constant sequence $(\bar a, \bar a, \bar a ,...)$ in $Q$, we have $\lim_{n\to \infty} \bar a = \bar a$.
Assumption b) is satisfied by any topology. (And by the way, assumption a) is equivalent to demanding that
a') The projection map $X \twoheadrightarrow Q$ is continuous.)
So here's your dilemma in a nutshell: Either you want to be able to conclude from convergence in $X$ to convergence of residues in $Q$, via assumption a). Then you have to live with the fact that in both your and my example, by virtue of the modded out subset being dense, this actually forces each sequence in $Q$ to converge to all points in Q simultaneously, i.e. the notion of convergence becomes totally meaningless.
But if you choose a topology on $Q$ in which it is a non-trivial statement to say that some sequence converges to some point (in particular if you want limits to be unique if they exist, i.e. the topology to be Hausdorff): then there will be sequences in $X$ which converge to a certain limit $L$ there, but whose projections in $Q$ will not converge to the projection of the limit $\bar L$.
I repeat, there are many (I guess uncontably many) topologies on $Q$ which are Hausdorff. But the above shows that for none of them, it is automatic that you can just the say the projection of a limit is the limit of the projections. I do not know if it is possible at all, and if it is, one needs to to put in significant work, to ensure that for all your sequences of interest $f^n(x)$ that property a) would still hold. This could be kind of an interesting question:
Is there a Hausdorff topology on $Q$ such that $\lim \tilde f^n(x\langle 2,3\rangle) = (x +2^{v_2(x)}3^{v_3(x)-1} )\cdot \langle 2, 3 \rangle$ for all $x \in X$, or at least for an interesting subset, like your $5$-rough integers.
