That's a very interesting problem. If one restricts to initial point $(x,1)$, its like a generalization of the question of points $(x,y) \in (\mathbb{F}_p)^2$ so that $xy\equiv 1 \bmod p$ with $(x,y)$ in certain regions of $[1,p]^2$. There's some relevant info on that in the comments at the question "Small residue class with small reciprocal" (Small residue classes with small reciprocal) where $x,y$ are both restricted. Based on that question you can get $(x,1)$ equivalent to a point in the box $[1,p^{3/4}]^2$ with $x \in [1,p^{3/4}]$. On the general question of points $(x,y)\in (\mathbb{F}_p)^2$ for curves satisfying certain conditions with $(x,y)$ in certain regions of $[1,p]^2$, there seems to be limitations imposed by lower bounds on "discrepancy" (after the curves are suitably normalized). I do not know about lower bounds on discrepancy, but there are upper bounds on discrepancy, by Granville, Shparlinski and Zaharescu:

A. Granville, I. E. Shparlinski, A. Zaharescu, On the Distribution of Rational Functions Along a Curve over $\mathbb{F}_p$ and Residue Races, J. Number Theory, vol 112 (2005), 216--237.

In your problem, if restricted to initial point $(x,1)$, you are asking if there is some $a \in [1,\sqrt{p}]$ so that there is a solution $(x,y)$ on the curve $xy\equiv a \bmod p$ where $x \in X=\mbox{set determined by the conditions you might want to specify}$ but $y \in [1,\sqrt{p}]$ so that $(x,1)$ is equivalent to $(a,y)$. For fixed $a$, if $|X|\sqrt{p}/p^2$ compares favourably with the upper bounds on discrepancy mentioned, I think you get a solution, and varying $a$ in the given range gives more freedom. Sorry if I've strayed a bit from your question!

That's a very interesting problem. If one restricts to initial point $(x,1)$, its like a generalization of the question of points $(x,y) \in (\mathbb{F}_p)^2$ so that $xy\equiv 1 \bmod p$ with $(x,y)$ in certain regions of $[1,p]^2$. There's some relevant info on that in the comments at the question "Small residue class with small reciprocal" (Small residue classes with small reciprocal) where $x,y$ are both restricted. Based on that question you can get $(x,1)$ equivalent to a point in the box $[1,p^{3/4}]^2$ with $x \in [1,p^{3/4}]$. On the general question of points $(x,y)\in (\mathbb{F}_p)^2$ for curves satisfying certain conditions with $(x,y)$ in certain regions of $[1,p]^2$, there seems to be limitations imposed by lower bounds on "discrepancy" (after the curves are suitably normalized). I do not know about lower bounds on discrepancy, but there are upper bounds on discrepancy, by Granville, Shparlinski and Zaharescu:

A. Granville, I. E. Shparlinski, A. Zaharescu, On the Distribution of Rational Functions Along a Curve over $\mathbb{F}_p$ and Residue Races, J. Number Theory, vol 112 (2005), 216--237.

In your problem, if restricted to initial point $(x,1)$, you are asking if there is some $a \in [1,\sqrt{p}]$ so that there is a solution $(x,y)$ on the curve $xy\equiv a \bmod p$ where $x \in X=\mbox{set determined by the conditions you might want to specify}$ but $y \in [1,\sqrt{p}]$ so that $(x,1)$ is equivalent to $(a,y)$. For fixed $a$, if $|X|\sqrt{p}/p^2$ compares favourably with the upper bounds on discrepancy mentioned, I think you get a solution, and varying $a$ in the given range gives more freedom. Sorry if I've strayed a bit from your question!