From the discussion above it looks like the answer is yes. Let the origin of your group be the point at infinity in the curve in $\mathbb{RP}^2$, and pick a topological group isomorphism of $S^1$ to the component of the identity to $S^1\cong \mathbb{R}/\mathbb{Z}$. The doublings of a point are given by truncating off the first $m$ digits of the base 2 expansion of your point. Thus the doublings of a point stay bounded if and only if the length of a consecutive string of 0's and 1's in this expansion is bounded above (there are plenty of irrational numbers with this property).

There's a similar answer for putting the origin somewhere else: you can never allow too much of the beginning of the expansion of the point at infinity to show up in the expansion of your point.

From the discussion above it looks like the answer is yes (EDIT: if you allow real numbers; the OP was unclear, perhaps they wanted a rational point, in which case I'm uncertain. Does anybody know anything about the binary expansion of complex numbers with rational Weierstrass p-values?). Let the origin of your group be the point at infinity in the curve in $\mathbb{RP}^2$, and pick a topological group isomorphism of $S^1$ to the component of the identity to $S^1\cong \mathbb{R}/\mathbb{Z}$. The doublings of a point are given by truncating off the first $m$ digits of the base 2 expansion of your point. Thus the doublings of a point stay bounded if and only if the length of a consecutive string of 0's and 1's in this expansion is bounded above (there are plenty of irrational numbers with this property).

There's a similar answer for putting the origin somewhere else: you can never allow too much of the beginning of the expansion of the point at infinity to show up in the expansion of your point.