"Is it possible for two spheres (a & b) to have an inelastic collision with BOTH the total linear and angular momentum preserved?"
More than that. It is not possible to have any collision in which they are not preserved.
However, you are not completely off-base here. Let's think about just what we mean by saying that "energy is not conserved" in a inelastic collision in the first place.
We don't actually mean that energy disappears, we mean that it disappears from the bulk kinetic terms (i.e. $\frac{1}{2} m_a v^2_{a1}$ and it's friends), and ends up in some other forms that we don't consider in our kinematics (mostly sound and heat for classroom demos).
Likewise some of the angular momentum could disappear from the bulk terms like $\vec{v}_{a1} \times m_a \vec{p}_{a1}$ into a channel that you are not writing down: the internal angular momentum of the product mass(es).
To be completely correct you should be attaching a moment of inertia and a angular velocity figure to each of your masses.
Next question for a simulation. When to almost-but-not-quite-point-masses with moments of inertia $I_{a,b}$ collide and stick what should the moment of inertia $I$ of the combined mass be and why?
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
Let's call our two colliding objects $A$ and $B$. So object $A$ and object $B$ come together, collide and ricochet away again. The collision may be elastic or inelastic.
You are quite correct that there must be a force acting during the collision, and because the force acts for some time there is an associated change in momentum. Consider just object $A$. If a force $F$ acts on $A$ for a time $t$ then the momentum of $A$ changes by the impulse $Ft$:
$$\Delta p_A = Ft$$
But remember that the force on $A$ is being exerted by $B$ during the collision. And Newton's third law tells us that the force being exerted on $A$ by $B$ must be equal and apposite to the force being exerted on $B$ by $A$. So if the force on $A$ is $F$, then the force on $B$ must be $-F$. Therefore the momentum change of $B$ is:
$$\Delta p_B = -Ft$$
The total momentum change is:
$$ \Delta p_{total} = \Delta p_A + \Delta p_B = Ft - Ft = 0 $$
and that's what conservation of momentum means. It means the total momentum is unchanged. The momenta of the individual objects $A$ and $B$ can and indeed do change, but the total momentum remains constant.