Let $\mathbb{H}=\{z\in\mathbb{C}:\Im(z)>0\}$ be the hyperbolic plane with the hyperbolic distance. Consider the tangent bundle T$\mathbb{H}=\mathbb{H}\times\mathbb{C}$ and the unit tangent bundle T$^1\mathbb{H}=\{(z,v)\in \text{T}\mathbb{H}:\|v\|_z=1\}$, with the norm induced by the inner product at $z$.
If we fix $z\in\mathbb{H}$ and $\mathbf{v}\in T^1_z\mathbb{H}$, then there is a unique geodesic that passes through $z$ in the direction of $\mathbf{v}$, and so we can define the geodesic flow $g_t:T^1\mathbb{H}\to T^1\mathbb{H}$ defined by following the uniquely defined geodesic for $(z,\mathbf{v})$ at time $t$. It can be shown that $g_t((i,i))=(e^ti,e^ti)$ and that $g_t((z,\mathbf{v}))=D(ga_t^{-1})(i,i)$, where $g$ is a Möbius transformation such that $g(i,i)=(z,\mathbf{v})$, $a_t=\left(\begin{matrix} e^{-t/2}& 0 \\ 0 & e^{t/2}\end{matrix}\right)$ and $Dg(z,v):=(g(z),g'(z)v)$.
I'd like to show that, given $(z,\mathbf{v})\in T^1\mathbb{H}$, then $$ \lim_{t\to\infty} d(g_t((i,i)),g_t(z,\mathbf{v})))=0\iff \Im(z)=1 \ \text{and } \mathbf{v}=i$$
The $\Leftarrow$ direction is easy, using the fact that $g_t((i,i))=(e^t i,e^ti)$ and $g_t((xì,i))=(x+e^ti,e^ti)$ and that the length of the horizontal path between this points if $\frac{|x|}{e^t}$.
I'm stuck in the converse. It seems geometrically obvious that $\mathbf{v}$ must be parallel to $i$, since otherwise the geodesic is a half-circle centered at the real axis, and hence the distance between the points does not tend to $0$. Thus we need only consider the case when the geodesic determined by $(z,\mathbf{v})$ is a vertical line, but that's as far as I got.
Let's consider $v=i$ first. Suppose $\operatorname{Im}z\ne 1$. Let $\zeta$ be the number with $\operatorname{Re}\zeta = \operatorname{Re}w $ and $\operatorname{Im}\zeta=1$. Then $$ d(g_t((i,i)),g_t((\zeta, i))) \to 0$$ as you have already shown. On the other hand, $d(g_t((z,i)),g_t((\zeta, i)))$ stays constant (and nonzero) because we are just moving both numbers along the same geodesic. By the triangle inequality, $ d(g_t((i,i)), g_t((z, i)))$ does not tend to zero.
Next, suppose $v\ne i$. Then $ d(g_t((z,i)), g_t((z, v)))\to \infty$, which is easiest to see in the disk model by placing $z$ in the center: two points move to the boundary along different radii. On the other hand, $ d(g_t((i,i)), g_t((z,i)))$ stays bounded, as one can check with a direct estimation of the distance between $e^t i$ and $\operatorname{Re}z + e^t\operatorname{Im} z$ (e.g., integrate $1/\Im\operatorname{zeta}$ along the line segment between them). Hence, $ d(g_t((i,i)), g_t((z, v)))\to \infty$.