Clarification regarding the molecular nature of friction described by Amonton's laws?

It is known that internal forces do no work. In this case, the frictional force does no work because it is internal. Hence, the kinetic energy of the block, in the beginning, must be equal to the final kinetic energy of the combined system, or in other words, no energy lost!

Total energy is (always) conserved, but part of the macroscopic kinetic energy of Block A is transferred to block B and the rest is converted to an increase in the microscopic internal kinetic energy of the two blocks due to internal kinetic friction work.

You didn't say how block A acquired its initial velocity on top of block B initially at rest, but let's just assume it was moving with horizontal velocity $V_1$ relative to the supporting surface when it gently landed on block B so that its initial velocity relative to block B and the supporting surface is $V_1$, and final velocity relative to block B is zero due to the friction force bringing it to a stop. Thereafter, both blocks move together at a final velocity $V_2$ on the supporting surface.

Now, looking at blocks A and B as a system, and considering the supporting surface to be frictionless, then there are no external forces (other than gravity) acting on the block AB system (the kinetic friction forces between the blocks being an internal forces of the system). Given that, the momentum of the system must be conserved.

The initial momentum of the system is that of block A, or $M_{A}V_1$, and the final momentum after block A comes to rest on block B is the momentum of the combined blocks, or $(M_{A}+M_{B})V_{2}$, where $V_1$ is the initial velocity of block A and $V_2$ is the final velocity of the combined blocks. For conservation of momentum we have

$$M_{A}V_{1}=(M_{A}+M_{B})V_2$$

$$V_{2}=frac{M_A}{(M_{A}+M_{B})}V_1$$

The initial kinetic energy of the system is the initial kinetic energy of block A, or

$$KE_{1}=frac{M_{A}V_{1}^2}{2}$$

The final kinetic energy of the system is

$$KE_{2}=frac{(M_{A}+M_{B})V_{2}^2}{2}$$

Substituting the second equation for $V_2$

$$KE_{2}=frac{M_{A}V_{1}^2}{2(1+frac{M_B}{M_A})}$$

We see that $KE_{2}<KE_1$, i.e., the final kinetic energy of the system is less than the initial kinetic energy. The "lost" kinetic energy being due to the internal friction work between the blocks. If the masses of the two blocks are equal, the final kinetic energy is half the initial.

In effect, we have a perfectly inelastic collision between the blocks where momentum is conserved but macroscopic kinetic energy is not. It is partially converted into an increase in the microscopic internal kinetic energy of the blocks (increase in temperature at the interface).

Hope this helps.