Haar measure - a problem from Folland

I was presented with this question from Folland's real analysis second edition involving Haar measures. It is problem 3 of chapter 11 page 347, which reads as follows:

Let G be a locally compact group that is homeomorphic to an open subset U of $ \mathbb{R}^n $ in such a way that, if we identify G with U, left translation is an affine map - that is, $ xy = A_x(y) + b_x $ where $ A_x $ is a linear transformation of $ \mathbb{R}^n $ and $ b_x \in \mathbb{R}^n $. Then $ det| A_x |^{-1} dx $ is a left Haar measure on G, where dx denotes Lebesgue measure on $ \mathbb{R}^n $. (Similarly for right translations and right Haar measures.)

I should mention we just got to Haar measures and topological groups in my class so it has not fully sunk in, and I have no idea how to do this. Thanks all helpers

Our potential left Haar measure $\mu$ on $G$ is given by (where $E$ is an element of the Borel sigma algebra of $G$): $$\mu(E) = \int_E \text{det}|A_y|^{-1} dy$$ So locally the volume of a group element $y \in G$ is scaled by the determinant of its inverse.

We want to show that the $\mu$ is left invariant with respect to the action of each affine map $A_x(y) + b_x$ associated to elements $x \in G$. In other words: $$\mu(A_x(E) + b_x) = \mu(E)$$ where $E$ is a set transformed by the map $A_x$ and translated by $b_x$.

First a note on the representation $A_x, b_x$. For two group elements $x, y$ we have that $x \mapsto A_x + b_x$ and $y \mapsto A_y + b_y$. We want to ensure that $xy$ maps to the composition of $x$ and $y$ so we must have when acting on a point $z$ that: $$A_x(A_y(z) + b_y) + b_x = A_{xy}(z) + b_{xy}$$

i.e. $A_{xy} = A_xA_y$ and $b_{xy} = A_xb_y + b_x$

We calculate: $$\mu(A_x(E) + b_x) = \int_{A_x(E) + b_x} \text{det}|A_y|^{-1} dy$$ Substituting $y = A_x(z) + b_x$ we have $$\mu(A_x(E) + b_x) = \int_{E} \text{det}|A_{A_x(z) + b_x}|^{-1} d(A_x(z) + b_x)$$ Since Lebesgue measure is known to be translation invariant and by scaling properties through linear maps: $$d(A_x(z) + b_x) = \text{det}|A_x|dz$$

The confusing looking integrand means that following, the linear component $A$ when multiplying the group element $x$ by $z$, i.e. $A_xA_z$. So we can rewrite this as: $$\mu(A_x(E) + b_x) = \int_{E} \text{det}|A_xA_z|^{-1} \text{det}|A_x|dx = \int_{E} \text{det}|A_z|^{-1} dx = \mu(E)$$

The other properties required to be a Haar measure follow from $dx$ being Lebesgue measure.