Fleshing out my comment slightly, here's a heuristic argument with a couple of blanks to be filled in that suggests that most points will be 'found' with probability $\lt 1$. For approximation purposes I'm going to ignore logarithmic factors here, so take all $O()$, etc. as being 'up to $n^{o(1)}$ factors'.

The argument is straightforward: pick some $P$ not in the initial assembly (e.g., $P=(0, 100)$.) Suppose that the asymptotic number of points in the construction after $n$ stages is $\Theta(n^\alpha)$ for some $1\leq \alpha \leq 2$. Then we have $\Theta(n^{2\alpha})$ pairs of points to choose; further suppose that $\Theta(n^\beta)$ of those pairs add the point $P$. Then at each iteration, $P$ gets added with probability $\Theta(n^{\beta-2\alpha})$ and therefore gets added at all with probability $\prod_n\left(1-\Theta(n^{\beta-2\alpha})\right)$; if we have $\beta\lt2\alpha-1$ then this product converges to a non-zero value.

From a different angle, note that if the asymptotic size is $\Theta(n^\alpha)$ then at each stage $\Theta(n^{\alpha-1})$ points are added. If there are $\Theta(n^\gamma)$ 'new' points accessible from the construction at stage $n$ and we assume that each point is added with equal probability, then any given accessible point will be added with probability $\Theta(n^{-(1+(\gamma-\alpha))})$; in particular, if it can be shown that the number of points immediately constructible from a figure of size $n^\alpha$ is $\Omega(n^{\alpha+\epsilon})$ for some $\epsilon\gt 0$ then we get the same convergence result. This seems likely to me if $\alpha\gt 1$, but of course filling in the blanks here is bound to be highly non-trivial.

Fleshing out my comment slightly, here's a heuristic argument with a couple of blanks to be filled in that suggests that most points will be 'found' with probability $\lt 1$. For approximation purposes I'm going to ignore logarithmic factors here, so take all $O()$, etc. as being 'up to $n^{o(1)}$ factors'.

The argument is straightforward: pick some $P$ not in the initial assembly (e.g., $P=(0, 100)$.) Suppose that the asymptotic number of points in the construction after $n$ stages is $\Theta(n^\alpha)$ for some $1\leq \alpha \leq 2$. Then we have $\Theta(n^{2\alpha})$ pairs of points to choose; further suppose that $\Theta(n^\beta)$ of those pairs add the point $P$. Then at each iteration, $P$ gets added with probability $\Theta(n^{\beta-2\alpha})$ and therefore never gets added at all with probability $\prod_n\left(1-\Theta(n^{\beta-2\alpha})\right)$; if we have $\beta\lt2\alpha-1$ then this product converges to a non-zero value.

From a different angle, note that if the asymptotic size is $\Theta(n^\alpha)$ then at each stage $\Theta(n^{\alpha-1})$ points are added. If there are $\Theta(n^\gamma)$ 'new' points accessible from the construction at stage $n$ and we assume that each point is added with equal probability, then any given accessible point will be added with probability $\Theta(n^{-(1+(\gamma-\alpha))})$; in particular, if it can be shown that the number of points immediately constructible from a figure of size $n^\alpha$ is $\Omega(n^{\alpha+\epsilon})$ for some $\epsilon\gt 0$ then we get the same convergence result. This seems likely to me if $\alpha\gt 1$, but of course filling in the blanks here is bound to be highly non-trivial.