I think all of the existing answers miss the real difference between energy and momentum in an inelastic collision.
We know energy is always conserved and momentum is always conserved so how is it that there can be a difference in an inelastic collision?
It comes down to the fact that momentum is a vector and energy is a scalar.
Imagine for a moment there is a "low energy" ball traveling to the right. The individual molecules in that ball all have some energy and momentum associated with them:
The momentum of this ball is the sum of the momentum vectors of each molecule in the ball. The net sum is a momentum pointing to the right. You can see the molecules in the ball are all relatively low energy because they have a short tail.
Now after a "simplified single ball" inelastic collision here is the same ball:
As you can see, each molecule now has a different momentum and energy but the sum of all of their momentums is still the same value to the right.
Even if the individual moment of every molecule in the ball is increased in the collision, the net sum of all of their momentum vectors doesn't have to increase.
Because energy isn't a vector, increasing the kinetic energy of molecules increases the total energy of the system.
This is why you can convert kinetic energy of the whole ball to other forms of energy (like heat) but you can't convert the net momentum of the ball to anything else.
The reason your expectations of kinetic energy loss are violated is because you picked a non-inertial reference frame.
In fact, you didn't notice, but based on your assumptions, momentum isn't even conserved.
Let's look more closely at your first equation:
$$m_e * 0 + m_i v_i = m_e * 0 + m_f v_f$$
Let's say the ball's mass is 5 kg. If the ball is falling down towards the ground, the initial velocity before the collision should be negative (assuming we have adopted a coordinate system where "up" is positive). Let's say the ball was moving at 10 m/s. Here is our equation so far:
$$m_e * 0 + (5 kg) (-10 m/s) = m_e * 0 + (5 kg) v_f$$
Let's simplify, given your assumption that the Earth doesn't move:
$$(5 kg) (-10 m/s) = (5 kg) v_f$$
Which gives us
$$ v_f = -10 m/s $$
Whaaa? The final velocity is negative? That means . . . after this "collision", the ball is still falling downward at the same speed! If we assume, as you did, that the velocity was positive, then the momentum of the system can't be conserved! Clearly, something is wrong here.
Here's the trouble: You began in the reference frame of the Earth. Once the ball hit the Earth, the Earth's reference frame is no longer inertial! You either have to introduce a fictitious force on the ball to account for the fact that the Earth accelerated (a tiny bit, yes, but an important tiny bit), or you need to find an inertial reference frame.
So, Let's pick the frame of reference in which the Earth starts out at rest. When the Earth moves, we'll let our frame of reference stay where it is, so as to allow it to remain inertial:
$$m_e * 0 + (5 kg) (-10 m/s) = m_e * v_{e, f} + (5 kg) v_f$$
$$-50 kg m/s = m_e * v_{e, f} + (5 kg) v_f$$
NOW we can introduce a coefficient of elasticity and demand that energy be either conserved or not, and find the final velocity of the Earth and the ball as they rebound from each other.
Note that although $v_{e, f}$ will be incredibly tiny, due to the enormous mass of the Earth $m_e$, the final momentum of the Earth will be non-negligible - in fact, the final momentum of the Earth must be comparable to the final momentum of the ball in order for momentum to be conserved!
So, to sum up: You picked a non-inertial reference frame, and didn't account for it, so your reliance on the laws of physics was betrayed.
Best Answer
A simple counterexample:
Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero.
Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not.
The crucial point is that kinetic energy depends on the square of velocity, $E_\mathrm{kin} = \frac{1}{2}mv^2$, and so is always positive - it cannot "cancel out" as momentum does, so momentum $\vec p = m \vec v$ can perfectly be conserved while the kinetic energy changes if the terms with "positive" and "negative" sign decrease or increase in a fashion that keeps the total momentum constant.