Let $G:=\text{SL}(2,\mathbb R)$, $\Gamma:=\text{SL}(2,\mathbb Z)$ and $u_t=\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}, t\ge 0$
Although this paragraph is not really needed to answer this question, I want to point out that ergodic theory tells us that for almost every point $g\Gamma$ in $G/\Gamma$ (with respect to Haar measure), we have $\{u_tg\Gamma\}_{t\ge 0}$ is dense in $G/\Gamma$ (Howe-Moore theorem tells us $u_t$ acts ergodically on $G/\Gamma$)
But how to produce any specific $g$'s such that $\{u_tg\Gamma\}_{t\ge 0}$ is dense in $G/\Gamma$? I guess there are some Diophantine approximation theorem that I am not familiar with. Perhaps we need $g$'s entries to be linearly independent over rationals.
If you want a specific matrix $g$, you can take $$ g=\left[ \begin{array}{cc} 1&0\\ \sqrt{2}&1\end{array}\right]. $$ Here is the reason. Let $U_\infty$ denote the subgroup of strictly upper triangular matrices in $G$. Then for $g\in G$, $$ U_\infty g = g U, \quad U=g^{-1}U_\infty g. $$ The unique fixed-point of $U$ in the ideal boundary of the hyperbolic plane is $g^{-1}(\infty)=\frac{a}{c}$, where $$ g^{-1}=\left[ \begin{array}{cc} a&b\\ c&d\end{array}\right]. $$ Thus, $U\cap \Gamma$ is a lattice in $U$ if and only if $g^{-1}(\infty)\in {\mathbb Q}\cup\{\infty\}$. In particular, for my choice of $g$ as above, $g^{-1}(\infty)= -1/\sqrt{2}\notin {\mathbb Q}$, hence, $U\cap \Gamma$ is not a lattice in $U$. Ratner proves (see here for a free copy of her paper) in
Ratner, Marina, Raghunathan’s conjectures for SL(2,R), Isr. J. Math. 80, No. 1-2, 1-31 (1992). ZBL0785.22013.
that for a nontrivial unipotent subgroup $U< G$ a coset $U\Gamma$ is either dense in $G/\Gamma$ or $U\cap \Gamma$ is a lattice in $U$. But, in fact, if you look at her proof closer, she proves density not only for the entire $U$-coset but also for cosets of connected semigroups in $U$. Thus, for $g$ as above, the $u_t$-semigroup orbit of $g\Gamma$ is dense in $G$.