Proving $dx.dy = r.drd\theta$ using x and y

If you want to do things algebraically then you'll have to resort to the exterior product:

$$\Bbb d x \wedge \Bbb d y = (\cos \theta \ \Bbb d r - r \sin \theta \ \Bbb d \theta) \wedge (\sin \theta \ \Bbb d r + r \cos \theta \ \Bbb d \theta) = \\ \cos \theta \sin \theta \ \Bbb d r \wedge \Bbb d r + r \cos^2 \theta \ \Bbb d r \wedge \Bbb d \theta - r \sin^2 \theta \ \Bbb d \theta \wedge \Bbb d r - r^2 \sin \theta \cos \theta \ \Bbb d \theta \wedge \Bbb d \theta$$

and, since $\Bbb d r \wedge \Bbb d r = \Bbb d \theta \wedge \Bbb d \theta = 0$ and $\Bbb d r \wedge \Bbb d \theta = - \Bbb d \theta \wedge \Bbb d r$ because the exterior product is graded-commutative (in particular anti-commutative for $1$-forms), we get

$$\Bbb d x \wedge \Bbb d y = r \ \Bbb d r \wedge \Bbb d \theta .$$

You can think about $dxdy$ as a small patch of area that is the parallelogram defined by the vectors $dx$ and $dy$. The key is that the area has a sign; so if we switch the order of $dx$ and $dy$, we get $dydx = -dxdy$. (One way to think about it is that the multiplication operator between $dx$ and $dy$ behaves like a cross product; it is anti-commutative.) Thus $drdr = -drdr$ so $drdr = 0$, similarly $d\theta d\theta = 0$.

Thus $$dxdy = (\cos(\theta)dr-r\sin(\theta)d\theta)(\sin(\theta)dr+r\cos(\theta)d\theta)\\=\cos(\theta)\sin(\theta)drdr-r\sin^2(\theta)d\theta dr+r\cos^2(\theta)drd\theta -r^2\sin(\theta)\cos(\theta)d\theta d\theta\\=0+r\sin^2(\theta)drd\theta +r\cos^2(\theta)drd\theta+0\\=rdrd\theta.$$ (The key is changing the $d\theta dr$ to $-drd\theta$ on the third line.)