Does the elasticity of a collision depend partially on the initial speed

The coefficient of restitution denoted by $e$ is the ratio of the final to initial relative velocity between two objects after they collide. In other words $$e=frac{v_f}{v_i}$$

And a per this Wiki article, if

$e = 0$: This is a perfectly inelastic collision. This means kinetic energy along the common normal is 0. Kinetic energy is converted to heat or work done in deforming the objects.

$0 < e < 1$: This is a real-world inelastic collision, in which some kinetic energy is dissipated.

$e = 1$: This is a perfectly elastic collision, in which no kinetic energy is dissipated, and the objects rebound from one another with the same relative speed with which they approached.

This means if $v_f =v_i$ we have a perfectly elastic collision and $e=1$.

If I now increase the initial velocity, and therefore the initial KE, I know that the KE right after impact would naturally also be higher than before.

Sure. If we have a collision where the final velocity is say $90 m/s$ and the intial velocity was $100 m/s$, then we have an "elasticity" (as you refer to it) or coefficient of restitution, $e=0.90$. If we now increase the initial velocity to say $111m/s$ then the final velocity will be about $100m/s$ the initial velocity, then $e$ is still $0.90$.

However, would it be proportional and the elasticity percent remain the same or not?

As you can see, the coefficient of restitution is a property of the material making up the object, and does not appear to be a property of initial velocity. But of course, as is usually the case in physics, we are dealing with ideal situations, and the question is valid. In fact, we should really be asking

Can the coefficient of restitution - COR - itself change if velocities are changed, specifically if velocities are increased?

As stated, $e$ is a property of the material, so can the properties of the material change at higher velocities?

Intuitively I feel like the higher the initial velocity the lower the elasticity percent but I don't have any science to back this up.

There are probably two major reasons why velocity would change the properties, and therefore the value of $e$, for a material.

First, we need to look at processes called deformation and whether this deformation is permanent or not. As we know, when a force is applied to an object, it will usually experience a change in its volume or overall shape, and if the deformation is elastic, once the force is removed the object will return to its original volume and shape. This is called an elastic deformation which is what we were talking about up to now. We describe this deformation using stress, strain and Young's modulus where $$Y=frac{sigma}{epsilon}$$ where $sigma$ is the stress and $epsilon$ is the strain and $Y$ is Young's modulus.

For relatively small values of $epsilon$, the body will return to its original shape and volume after the stress is removed. This is called elastic strain. But if the strain is too large in comparison, then the body will plastically deform or even fracture. As oppose to elastic strain, plastic strain permanently alters the shape and elastic characteristics of the object, which leads us to how $e$ can change as a function of $v$.

The other reason why there would be a velocity dependence, is due to temperature affects. The elasticity of objects is affected by temperature, and the temperature of a material changes due to collisions. The temperature increases due to an increase in the internal energy brought about by the deformation during the collision. Therefore the value of $e$ once again should change due to changing values of $v$.

It appears your original intuition is indeed correct here, since high velocity impacts can indeed change (permanently or non-permanently) the properties of a material when it collides. The following diagram is obtained from this study

which clearly shows how $e$ decreases when $v$ is increased. We can investigate this by looking at objects dropped from a height, knowing that the COR is given by $$e=frac{v_f}{v_i}=sqrt{frac{h_f}{h_i}}$$ where $h_i$ is the drop height and $h_f$ is the rebound height. We can see that these heights are proportional to the squares of the velocities and a study here where they look at bouncing balls (non peer reviewed), finds that as velocity (higher drop height) increases, $e$ does decrease as you suspected. And in fact, they find that $$e=0.88h_i^{-0.1}$$

Another paper here goes into this subject in more depth and models a more complicated relationship between $e$ and velocity.

Another study here finds another relationship between $v_i$ and $e$ such that $$e=cv_i^{-a}$$ for certain constants $c$ and $a$. This seems to show an inverse relationship (assuming $age 1$) between $e$ and $v_i$ which is consistent again with your earlier assumption.