= Solution
Let $v=(x,y)^T$ and $B(\rho)=\left(\begin{smallmatrix}\cosh\rho&\sinh\rho\\\sinh\rho&\cosh\rho\end{smallmatrix}\right)$. Matrix multiplication gives
$$
(\rho,v)(\rho',v')=(\rho+\rho',v+B(\rho)v').
$$
Thus $G$ is a <semidirect product> of the translation plane by the one-parameter <Lorentz boost> group. Differentiating left multiplication at the identity gives the <left-invariant vector fields>
$$
\boxed{X_0=\partial_\rho,\qquad
X_1=\cosh\rho\,\partial_x+\sinh\rho\,\partial_y,\qquad
X_2=\sinh\rho\,\partial_x+\cosh\rho\,\partial_y.}
$$
They obey $[X_0,X_1]=X_2$, $[X_0,X_2]=X_1$ and $[X_1,X_2]=0$.
Their dual <coframe> is $d\rho$, $\cosh\rho\,dx-\sinh\rho\,dy$, and $-\sinh\rho\,dx+\cosh\rho\,dy$. The <Haar measure from a left-invariant coframe> is their absolute wedge product. Since $\det B(\rho)=1$,
$$
\boxed{d\mu=C\,d\rho\,dx\,dy,\qquad C>0.}
$$
One can also check this directly: left multiplication has Jacobian determinant one. Right multiplication has a block triangular Jacobian with determinant one too, so the same measure is both left and right <Haar measure>.
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