A Haar measure via the Lebesgue measure on $\Bbb R^d$ $\newcommand{\d}{\mathrm{d}}$
This the Exercise 3, Chapter 11 of the Gerald B. Folland book Real Analysis:

Let $G$ be a locally compact group that is homeomorphic to an open subset $U$ of $\Bbb R^d$ 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 $\Bbb R^d$ and $b_x\in\Bbb R^d$. Then $|\det A_x|^{-1}\d x$ is a left Haar measure on $G$, where $\d x$ denotes Lebesgue measure on $\Bbb R^d$.

Let me know if I understand this.
What the problem gives us is $G$ a locally compact group, an open set $U\subseteq\Bbb R^d$, and a bijection $\varphi:G\to U$ such that $\varphi$ and $\varphi^{-1}$ are both continuous and satisfy that given $u\in G$ there is a linear operator $A_{\varphi(u)}:\Bbb R^d\to \Bbb R^d$ and $b_{\varphi(u)}\in\Bbb R^d$ so that $$\varphi(uv)=A_{\varphi(u)}(\varphi(v))+b_{\varphi(u)}.$$

Is this right

If it is, define $f:\Bbb R^d\to\Bbb [0,\infty[$ given by $$f(x)=|\det A_x|^{-1}\quad\text{i.e.}\quad f(x)=|\det A_{\varphi(\varphi^{-1}(x))}|^{-1}.$$

Is the problem asking if the measure $\mu$ in $G$ given by $$\mu(E)=\int_{\varphi(E)}f(x)\d x$$ is a left Haar measure?

This reminds me the formula $$\int_E f(y)\d y=|\det T|\int_{T^{-1} E} f(Tx)\d x,$$
but I don't know what  can I do.
Any advice in how to interpret the problem or on how to proceed is very appreciated.
 A: The notation $xy=A_xy+b_x$ is quite confusing indeed.  It appears on the left hand side, that $x,y\in G$; however, on the right hand side, it would appear that $y\in\mathbb{R}^d$. Your introduction of the homeomorphism, $\varphi:G\mapsto U\subset\mathbb{R}^d$, is the right thing to do.

Is this right

Yes, that is a restatement of part of the hypothesis with the homeomorphism given explicitly by $\varphi$. Since $\varphi$ is a homeomorphism, we could simply write
$$
\varphi(uv)=A_u\varphi(v)+b_u
$$

Is the problem asking if the measure $\mu$ in $G$ given by $$\mu(E)=\int_{\varphi(E)}f(x)\mathrm{d} x$$ is a left Haar measure?

Since $f(x)$ is meaningless for $x\not\in U$, I would define $f:U\mapsto[0,\infty)$ by
$$
f(\varphi(u))=|\det A_u|^{-1}
$$
and then define $\mu$ as you do above:
$$
\begin{align}
\mu(E)
&=\int_{\varphi(E)}f(x)\,\mathrm{d}x
\end{align}
$$
Now note that
$$
\begin{align}
\varphi(uvw)
&=\color{#C00000}{A_{uv}}\varphi(w)+\color{#00A000}{b_{uv}}\\
&=A_u\varphi(vw)+b_u\\
&=A_u(A_v\varphi(w)+b_v)+b_u\\
&=\color{#C00000}{A_uA_v}\varphi(w)+\color{#00A000}{A_ub_v+b_u}
\end{align}
$$
Therefore, we get $A_{uv}=A_uA_v$. Thus, we get
$$
\begin{align}
f(\varphi(uv))
&=|\det(A_{uv})|^{-1}\\
&=|\det(A_uA_v)|^{-1}\\
&=|\det(A_u)\det(A_v)|^{-1}\\
&=f(\varphi(u))f(\varphi(v))
\end{align}
$$
At this point, to show the translation invariance,
$$
\begin{align}
\mu(uE)
&=\int_{\varphi(uE)}f(x)\,\mathrm{d}x\\
&=\int_{\varphi(E)}\color{#C00000}{f(\varphi(u\varphi^{-1}(x)))}\,\color{#00A000}{\mathrm{d}(A_{u}x+b_{u^{-1}})}\\
&=\int_{\varphi(E)}\color{#C00000}{f(\varphi(u))f(x)}\color{#00A000}{|\det A_{u}|\,\mathrm{d}x}\\
&=\int_{\varphi(E)}|\det A_{u}|^{-1}f(x)|\det A_{u}|\,\mathrm{d}x\\
&=\int_{\varphi(E)}f(x)\,\mathrm{d}x\\
&=\mu(E)
\end{align}
$$
