I did not realised that the structure of this group would be as simple as it turned out to be, mostly because I had simply assumed that working with this group would be beyond me given that I've never worked with Lie groups before. After looking again, I believe I've proved that G' is a Lie group and found an atlas for it.
The group G' is isomorphic to the group of real $2\times 2$ matrices with nonzero determinant of the form $\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}$ as the equivalence class of $\begin{pmatrix} x & 0 \\ y & z \end{pmatrix}$ contains precisely $1$ matrix of such form, specifically $\begin{pmatrix} \frac{x}{z} & 0 \\ \frac{y}{z} & 1 \end{pmatrix}$ (note that $z\neq 0$ because $xz \neq 0$).
G' can be made into a topological space by equipping it with the metric defined as d\left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix},\begin{pmatrix} x' & 0 \\ y' & 1 \end{pmatrix}\right) = \sqrt{(x-x')^2 + (y-y')^2}. An atlas for G' is given by the charts \alpha: G'_+ \rightarrow \mathbb{R}^2 and \beta: G'_- \rightarrow \mathbb{R}^2, where G'_+ is the (open) set of matrices with positive determinant and G'_- the (open) set with negative determinant, defined by $\alpha \left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}\right) = \begin{pmatrix} x \\ y \end{pmatrix}$ and $\beta \left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}\right) = \begin{pmatrix} x \\ y \end{pmatrix}$. These charts are non-intersecting, and this makes G' a smooth manifold.
In order to show that G' is a Lie group we need only show that the map \phi: G'\times G' \rightarrow G' defined as \phi \left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix},\begin{pmatrix} x' & 0 \\ y' & 1 \end{pmatrix},\right) = {\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}}^{-1} \begin{pmatrix} x' & 0 \\ y' & 1 \end{pmatrix} = \begin{pmatrix} \frac{x'}{x} & 0 \\ \frac{xy'-yx'}{x} & 1 \end{pmatrix} is smooth. This follows easily from the fact that rational functions on $\mathbb{R}^2$ are smooth whereever their denominator is not $0$ and that $x\neq 0$ in G'.