How do I prove that two equations in Cartesian and Polar coordinates are equivalent?

In $\mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way: $$\begin{cases} x = \rho \cos \theta \\ y = \rho \sin \theta. \end{cases} $$ Of course I need to specify more about $\rho$ and $\theta$, for now $\rho \geq 0$ and $\theta \in [0,2\pi)$. Note that $$\begin{cases} \mathrm{d}x = \cos \theta \,\mathrm{d}\rho -\rho\sin \theta \,\mathrm{d}\theta \\ \mathrm{d}y = \sin \theta \,\mathrm{d}\rho + \rho \cos \theta \,\mathrm{d}\theta, \end{cases}$$ hence \begin{align} \mathrm{d}x \wedge \mathrm{d}y & = (\cos \theta \,\mathrm{d}\rho -\rho\sin \theta \,\mathrm{d}\theta) \wedge (\sin \theta \,\mathrm{d}\rho + \rho \cos \theta \,\mathrm{d}\theta) \\ & = \rho \,\mathrm{d}\rho \wedge \mathrm{d}\theta. \end{align} Therefore the transformation is singular when $\rho = 0$. Also, reversing the equations above you get $$ \begin{cases} \rho^2 = x^2+y^2 \\ \tan \theta = y/x, \end{cases} $$ so when $\theta = \pi/2+k\pi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $\rho > 0$ and $\theta \neq \pi/2 +k\pi$. This way the coordinate transformation is invertible.

Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $ becomes \begin{align} (\rho \cos \theta-1)^2+\rho^2\sin^2\theta -1 & = 0 \\ \rho^2-2\rho\cos \theta& = 0 \\ \rho(\rho-2\cos \theta)&=0, \end{align} which implies $\rho = 2\cos \theta$, because $\rho \neq 0$. Actually $\rho > 0$, which implies $\cos \theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $\rho = 0$ is also a solution of $\rho(\rho-2\cos \theta)=0$, thus just study this case on its own. It is trivial, because $\rho = 0$ corresponds to the origin $(x,y) = (0,0)$.

Note that your original circle (in Cartesian coordinates) is a circle centered at $(1,0)$ of radius $1$. So the entire circle is in quadrants 1 and 4. So $\theta$ must be in the interval $[-\pi/2,\pi/2]$. Otherwise you just go over the circle again and again. You could have also noticed that you were given $\cos\theta\gt 0$, so it will also restrain your solutions.

The way I tend to approach this problem is to explicitly use $x=r\cos\theta$ and $y=r\sin\theta$ in the first equation on the left hand side. If you get $1$, then the proof is done.