Question: Consider the map from the title $$f \colon \text{SL}(2,\mathbb{R}) \to \mathbb{R}^2 \setminus \{0\},\qquad A \mapsto A\begin{pmatrix}1 \\ 0 \end{pmatrix}.$$
Is it a closed map, i.e. does it map closed sets $A \subseteq \text{SL}(2,\mathbb{R})$ to closed sets $f(A) \subseteq \mathbb{R}^2 \setminus \{0\}$?
What I've tried: According to Wikipedia, this is equivalent to the condition that for all subsets $A \subseteq \text{SL}(2,\mathbb{R})$, we have $$ \overline{f(A)} \subseteq f(\overline{A}). $$
So, for example, if I take the sequence of matrices $$ a_n =\begin{pmatrix} 2 & n \\ 1/n & 1 \end{pmatrix}, $$ then we have $f(a_n) \to \begin{pmatrix}2 \\ 0 \end{pmatrix}$ but $(a_n)_n$ itself does not converge. So maybe one can construct a set $A$ containing the image of this sequence, but its closure does not contain a matrix with first row $\begin{pmatrix}2 \\ 0 \end{pmatrix}$?
You are nearly there: $$A=\left\{\pmatrix{2&b\\c&1}:bc=1\right\}$$ is a closed subset of $SL_2(\Bbb R)$ but $f(A)$ is not closed, by your example.