How do you find the isometries of a manifold $M$ with metric $ds^2=\frac{du^2}{u^2}+\frac{dv^2}{v^2}?$
Observe that $ds^2$ is of the form $ds^2=g(u)du^2+f(v)dv^2.$ This means that the metric must describe a Euclidean geometry. In fact with more work we can show that $ds^2$ is a transported metric from the familiar $\Bbb R^2$ to $M:=\Bbb R^2_+$ via $\exp.$ I calculated the pullback via an immersion here:
\begin{align} (f^{-1})^* (dx^2 + dy^2) &=( d((f^{-1})^*x))^2 + ( d((f^{-1})^*y))^2 \\ &= (d (x\circ f^{-1}))^2 +(d (y\circ f^{-1}))^2 \\ &= (d \log u)^2 +(d\log v)^2 \\ &= \left( \frac{1}{u} du\right)^2 + \left( \frac{1}{v} dv\right)^2 \\ &= \frac{1}{u^2} du^2 + \frac{1}{v^2} dv^2. \end{align}
Now I know that the isometries of $M$ are basically the isometries of $\Bbb R^2$ in disguise.
I figured out that translation in $\Bbb R^2 \implies (x+a,y+b)$ has analogue in $M \implies (ax,by).$
Rotation in $\Bbb R^2 \implies \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}(x,y)$ has what analogue in $M?$
I couldn't figure this one out.
You have shown that: \begin{align} f \colon (\mathbb{R}_+^*\times \mathbb{R}_+^*,g) &\longrightarrow (\mathbb{R}^2,\mathrm{can})\\ (u,v) & \longmapsto (\ln u, \ln v) \end{align} is an isometry. Let $\theta \in \mathbb{R}$ and $R_{\theta}$ be the corresponding rotation in $\mathbb{R}^2$. Then:
\begin{align}f^*R_{\theta}(u,v) &= f^{-1} \left(R_{\theta}f(u,v)\right)\\ &= f^{-1}\left(\cos \theta \ln u +\sin\theta \ln v, -\sin \theta \ln u + \cos \theta \ln v \right) \\ &= \left(\exp \left(\cos \theta \ln u +\sin\theta \ln v \right), \exp\left( -\sin \theta \ln u + \cos \theta \ln v\right) \right)\\ &= (u^{\cos \theta}v^{\sin \theta}, u^{-\sin\theta}v^{\cos\theta}). \end{align}
Also, translations $(x,y) \mapsto (x+a,y+b)$ become $(u,v)\mapsto (e^au,e^bv)$.