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.
Consider your example 1 with two billiard balls, mass $0.16 \,\rm kg$, colliding with a red ball falling down with a speed of $5\, \rm m\,s^{-1}$ and colliding with a stationary white ball.
Applying conservation of linear momentum (assuming there are no external force and the collision is elastic) results in the red ball momentarily stopping and the white ball moving downwards at $5\, \rm m\,s^{-1}$.
The impulse (change of momentum) on each ball is $0.16 \times 5 = 0.8 \, \rm N\,s$.
During the collision the gravitational force would have had a effect on the balls but to know how much one must know the collision time.
Peter Bohacek has produced many Direct Measurement Videos and the relevant one for this answer is Billiard Ball Collision three consecutive frames from which are shown below.
This shows that the collision time is less than $0.001 \, \rm s$.
Going back to the falling red and white billiard balls, in a time of $0.001 \, \rm s$ the impulse due to gravity on one of the balls is $\text{~}\, 0.16 \times 10 \times 0.001 = 0.016 \,\rm Ns $ which is very much smaller than the impulse on the balls due to the collision, $ 0.8 \, \rm N\,s$.
So the assumption of a very short collision time resulting in very little effect on the outcome of the collision is a good one.
More time could have been spent analysing the video to get a more accurate upper bound for the collision time which might be the basis of a nice assignment?
Best Answer
The change of momentum due to an applied force is given by the impulse of the interaction. This is true for any object in any interaction: $$J = F\Delta t$$ During a typical collision problem, both $F$ and $\Delta t$ are small compared to the force of collision and the duration of the collision. Hence, the impulse due to gravity can be neglected.