Relationship between angular momentum and centripetal force in Newtonian orbits

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.

In general, a bound orbit in Newtonian gravity is an ellipse where the radial distance varies with time and the velocity is not always perpendicular to the position. The formulae you are using, which only work for circular orbits, therefore do not apply when considering Earth's orbit as an ellipse. What you have shown is that circular orbits of different radii around a given central mass do not have the same specific angular momentum, and not anything about elliptical orbits. For a quick derivation of the general non-circular case, see my answer here.

The relation $a=v^2/r$ is for a circular orbit, leading to $v=\sqrt{GM/r}$ being the velocity of a circular orbit at radius $r$. Since you find that for a general orbit $v\propto 1/r$ at the highest and lowest points of the orbit, and this is a stronger scaling than the circular orbit $v\propto 1/\sqrt{r}$, you can conclude that the Earth moves faster than a circular orbit of the same radius ($v>\sqrt{GM/r}$) when it is at the lowest point in its orbit and slower than a circular orbit of the same radius ($v<\sqrt{GM/r}$) when it is at the highest point in its orbit.

(Note that your $v\propto 1/r$ is only valid at the highest and lowest points of the orbit, since in deriving it you assumed that there is no radial component to the motion; otherwise $L=mrv$ would not hold).

1. In this answer let us for simplicity only consider circular orbits. OP's eqs. (1) & (2) are closely related to Kepler's 2nd and 3rd law, respectively, and their coexistence.

2. The main point is that when we state a power law (or a law of proportionality), it is important to understand the limitations of the law, i.e. the assumptions that make the proportionality constant actually constant. (To see another example of what can go wrong, see e.g. this Phys.SE post.)

3. In OP's case: If we compare 2 different circular orbits, there is no reason why they should have the same angular momentum, i.e. in that comparison, we should discard OP's eq. (1).

Try using the Kepler's laws to figure this out this same result is obtained using the fact that areal velocity stays the same

dA/dt = constant also dA/dt = L/2M ( Equations used for Ellipses) 

L = mvr also using v = root over Gm/ r which is the orbital velocity

putting value of v we find the V is proportional to reciprocal of root over r

Try diving into Kepler's law it might clear your doubts up !