I'm trying to calculate the left invariant vector fields on $\mathbb{R}^2$ from the action $\mu: G \times \mathbb{R}^2 \to \mathbb{R}^2$ with the multiplication $$\mu((a,b,c), (x,y)) = (ax-cy, by). $$ I know the identity element on $e\in G$ is $e = (1,1,0)$, and I've calculated
$$ \begin{align*} \mu_{g\ast}^1(\partial_x) &= \partial_x(\mu_g^1) = a, \\ \mu_{g\ast}^2(\partial_x) &= \partial_x(\mu_g^2) = 0,\\ \mu_{g\ast}^1(\partial_y) &= \partial_y(\mu_g^1) = -c, \\ \mu_{g\ast}^2(\partial_y) &= \partial_y(\mu_g^2) = b, \end{align*} $$ giving $X_1= a\partial_x$ and $X_2 = b\partial_y - c\partial_x$ as the left invariant vector fields, but now I'm not sure how to write $a,b,c$ in terms of $x$ and $y$. Am I missing something?