Your problem is in equating

$$\frac{GMm}{r^2}=\frac{mv^2}{r},\quad\text{ or equivalently}\quad\frac{GM}{r^2}=\frac{v^2}{r}.$$

Indeed, since the centripetal acceleration is $a=GM/r^2$, you might expect to set this equal to $v^2/r$ which is equal to the centripetal acceleration in cases of circular motion. But the Earth isn't moving in a circle, it's moving in an ellipse.

These basic problems relating to orbital mechanics were solved by Newton long ago. Today one of the primary models of orbital velocity is the vis-viva equation,

$$v^2=GM\bigg(\frac{2}{r}-\frac{1}{a}\bigg)$$

for the semi-major axis $a$ of the orbit. Note also that $v$ here is the relative speed between the two bodies; since the Sun also sort of orbits around the Earth, there is a slight degree of error, but because the Sun is much bigger than the Earth (citation needed) it's pretty close all things considered.

Verifying that orbital angular momentum is conserved is an elementary exercise treated in many books, e.g. Fundamentals of Astrodynamics, and as such I will not repeat it here, but you can also find other questions on that topic on this site e.g. here.