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.
the change in momentum should be the vector sum of initial and final momentum
That's incorrect because you must subtract the quantities in order to obtain the difference between, i.e., by how much it changed.
That becomes clear when considering $\vec{p}=\text{const.}$, say, $\vec{p}=3\hat{\imath}$ in your units; then, according to the quoted text above, the change would be $\Delta \vec{p} = \vec{p}_i +\vec{p}_f = 3\hat{\imath}+3\hat{\imath} = 6\hat{\imath}$, which doesn't make sense, since the change of a constant quantity is zero (since constant means that it doesn't change), which is what you get from your second way of calculating: $$ \Delta \vec{p} = \vec{p}_f -\vec{p}_i.$$
Then, for the constant $\vec{p}$ example, you get $ \Delta \vec{p} = 3\hat{\imath}-3\hat{\imath} = \vec{0}$.
Best Answer
The two vehicles have separate momenta initially, and share a momentum after collision. While the magnitude of change in momentum for each individual vehicle can differ, the system conserves momentum as a whole.
Consider two objects colliding, one with momentum $p_1$ going right, and one going left with momentum $p_2 = -2p_1$. Before collision, it is clear that the magnitude of momentum of the second object is larger than the first, or $p_2 > p_1$.
After collision, the objects stick together and as a whole move with momentum $-p_1$, due to conservation of momentum.
So the change in $p_1$, $\Delta p_1 = -2 p_1 = p_2$
and the change in $p_2$, $\Delta p_2 = p_1$
Here, it is clear that the magnitude of change in momentum for $p_1$ is larger than the magnitude of change in momentum for $p_2$, or $|\Delta p_1| > | \Delta p_2 |$. But each momentum individually is changed by the opposing momentum.
But if we look at the momentum of the system, we see that the quantity is conserved by $p_1 + p_2 = p_1 - 2p_1 = -p_1$
This implies that although the momenta of the two objects change by different amounts, they change by amounts which will ensure that the total momentum after equals the sum of the momenta initially. $p_1$ is changed more than $p_2$, but the sum of $p_1$ & $p_2$ does not change.
Using conservation of momentum, we know the momentum before will equal the momentum of the system after, or
$p_1 + p_2 = p_{1}'+p_{2}'$
Let $p_1$ be car momentum, $p_2$ be truck.
$p_1 = mv_1$
$p_2 = -Mv_1$
After collision, the vehicles are stuck together, so they will share their mass and velocity.
$p_{1}'+p_{2}'= v_2(m+M)$
So the conservation of momentum yields: $mv_1 - Mv_1 = v_1(m-M) = v_2(m+M)$
The momentum change in the car is equal to the momentum of the truck, and conversely, the momentum change in the truck is equal to the car.
So, $\Delta p_1 = p_2$
and, $\Delta p_2 = p_1$
Then if $|p_2|>|p_1|$, $|\Delta p_1| > |\Delta p_2|$
The change of momentum for the car is more than the change of momentum for the truck, but the changes are equal and opposite leading to the conservation of momentum.