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And this is exactly what happens and what $f_s$ takes care of on the top book in your example. Had its mass been larger, then it would be tougher to induce the same acceleration in it as that of the bottom book. Then a larger $f_s$ would be needed to accelerate it just as much. But in the constant-velocity situation with no acceleration anymore, the $f_s$ doesn't have to worry about that - it only takes care of the air resistance which is unrelated to the mass.

See more details below.

Why they have the same velocity?

Because static friction is an adjustable force. Its aim is to prevent sliding, and it adjust to whatever is needed to prevent that.

• The forces acting on the bottom book are the push $P$, air resistance $D_{bottom}$, bottom kinetic friction $f_k$ and top static friction $f_s$, and they all cancel to zero, since we have no acceleration, according to Newton's 1st law: $$\text{Bottom book:}\quad \sum F=P-D_{bottom}-f_k-f_s=0$$

• The forces acting on the top book are air resistance $D_{top}$ and the static friction $f_s$, which is equal to that before but opposite: $$\text{Top book:}\quad \sum F=-D_{top}+f_s=0$$

As you can see for the top book, $f_s$ only has to counter the air resistance. And this doesn't depend on the top book's mass. So $f_s$ has got nothing to do with the mass here. The mass could be anything - double as much, 10 times as must, 100 times as much - and that wouldn't influence $f_s$. (Only if you changed the shape of the book, you would change the air resistance, and then $f_s$ would have to change accordingly to still counteract the air resistance exactly.)

Now, both of these equations are true at the same time in this constant-speed situation.

Now, imagine that you suddenly push double as hard:

1. $\sum F=P-D_{bottom}-f_k-f_s$ is then no longer zero, but equal to $ma$ (Newton's 2nd law), so the bottom book accelerates.
2. The top book follows along, since $f_s$ will adjust and increase to prevent sliding, so $\sum F=-D_{top}+f_s$ no longer equals zero, but $ma$. $f_s$ grows so it can counteract $D_{top}$ as well as induce the necessarily acceleration at the same time. And this $a$ equals that of the bottom book - otherwise there would be sliding.
3. This acceleration increases the velocity of both books, which increases $D_{top}$ and $D_{bottom}$, since they depend on velocity.
4. As they increase, they reduce the total force, and both $\sum F=P-D_{bottom}-f_k-f_s$ and $\sum F=-D_{top}+f_s$ will reduce and gradually become closer to zero. The air resistances in this way gradually balances out the equations again and the accelerations gradually decreases.
5. In turn, these decreasing accelerations cause $f_s$ to be smaller, since a smaller force is needed to induce this smaller acceleration in the top book.
6. Soon, $f_s$ and $D_{top}$ meet and become equal in $\sum F=-D_{top}+f_s$ so the total force on the top book equals zero again. This can only happen when also $\sum F=P-D_{bottom}-f_k-f_s$ has balanced out to zero again, because only then are there no more accelerations of the bottom book, which $f_s$ has to induce into the top book as well. $f_s$ is left to take care of only the air resistance on the top book, although at a new value.

So, the new push has caused a new velocity, new air resistances and a new $f_s$. $f_k$ has been constant all the way, since it doesn't depend on velocity or anything else that is varying here. All in all, the $f_s$ will always adjust to keep the books from sliding. It has a limit of course, and if it at anytime during these steps had to be larger than its limit, then it would let go and sliding would start (and kinetic friction would replace it).

And this is exactly what happens and what $f_s$ takes care of on the top book in your example. Had its mass been larger, then it would be tougher to induce the same acceleration in it as that of the bottom book. Then $f_s$ would increase the total force in order to accelerate it just as much. But in the constant-velocity situation with no acceleration anymore, the $f_s$ doesn't have to worry about that - it only takes care of the air resistance which is unrelated to the mass.

Why they have the same velocity?

So, the new push has caused a new velocity, new air resistances and a new $f_s$. $f_k$ has been constant all the way, since it doesn't depend on velocity or anything else that is varying here. All in all, the $f_s$ will always adjust to keep the books from sliding. It has a limit of course, and if it at anytime during these steps had to be larger than its limit, then it would let go and sliding would start (and kinetic friction would replace it).

The forces that act on them are different and their masses are different so the books have different acceleration

This is not correct. Remember again Newton's 2nd law $\sum F=ma$. If both the total forces on and the masses of two objects are different, then that doesn't imply that the accelerations are different. One could easily have double the force but half the mass and thus have the exact same acceleration as the other.

And this is exactly what happens and what $f_s$ takes care of on the top book in your example. Had its mass been larger, then it would be tougher to induce the same acceleration in it as that of the bottom book. Then a larger $f_s$ would be needed to accelerate it just as much. But in the constant-velocity situation with no acceleration anymore, the $f_s$ doesn't have to worry about that - it only takes care of the air resistance which is unrelated to the mass.

The forces that act on them are different and their masses are different so the books have different acceleration

This is wrong logic. Remember Newton's 2nd law $\sum F=ma$. If both the total forces on and the masses of two objects are different, then that doesn't imply that the accelerations are different. One could easily have double the force but half the mass and thus have the exact same acceleration as the other.

And this is exactly what happens and what $f_s$ takes care of on the top book in your example. Had its mass been larger, then it would be tougher to induce the same acceleration in it as that of the bottom book. Then a larger $f_s$ would be needed to accelerate it just as much. But in the constant-velocity situation with no acceleration anymore, the $f_s$ doesn't have to worry about that - it only takes care of the air resistance which is unrelated to the mass.

See more details below.

Why they have the same velocity?

So, the new push has caused a new velocity, new air resistances and a new $f_s$. $f_k$ has been constant all the way, since it doesn't depend on velocity or anything else that is varying here. All in all, the $f_s$ will always adjust to keep the books from sliding. It has a limit of course, and if it at anytime during these steps had to be larger than its limit, then it would let go and sliding would start (and kinetic friction would replace it).

1. $\sum F=P-D_{bottom}-f_k-f_s$ is then no longer zero, but equal to $ma$ (Newton's 2nd law), so the bottom book accelerates.
2. The top book follows along, since $f_s$ will adjust and increase to prevent sliding, so $\sum F=-D_{top}+f_s$ no longer equals zero, but $ma$. $f_s$ makes sure that this $a$ equals that of the bottom book - otherwise there would be sliding.
3. This acceleration increases the velocity of both books, which increases $D_{top}$ and $D_{bottom}$, since they depend on velocity.
4. As they increase, they reduce the total force, and both $\sum F=P-D_{bottom}5. f_k-f_s$ and $\sum F=-D_{top}+f_s$ will reduce and gradually become closer to zero. The air resistances in this way gradually balances out the equations again and the accelerations gradually decreases.
5. In turn, these decreasing accelerations cause $f_s$ to be smaller, since a smaller force is needed to prevent sliding.
6. Soon, $f_s$ and $D_{top}$ meet and become equal in $\sum F=-D_{top}+f_s$ so the total force on the top book equals zero again. This can only happen when also $\sum F=P-D_{bottom}-f_k-f_s$ has balanced out to zero again, because only then are there no more accelerations of the bottom book, which $f_s$ has to induce into the top book as well. $f_s$ is left to take care of only the air resistance on the top book, although at a new value.

1. $\sum F=P-D_{bottom}-f_k-f_s$ is then no longer zero, but equal to $ma$ (Newton's 2nd law), so the bottom book accelerates.
2. The top book follows along, since $f_s$ will adjust and increase to prevent sliding, so $\sum F=-D_{top}+f_s$ no longer equals zero, but $ma$. $f_s$ grows so it can counteract $D_{top}$ as well as induce the necessarily acceleration at the same time. And this $a$ equals that of the bottom book - otherwise there would be sliding.
3. This acceleration increases the velocity of both books, which increases $D_{top}$ and $D_{bottom}$, since they depend on velocity.
4. As they increase, they reduce the total force, and both $\sum F=P-D_{bottom}-f_k-f_s$ and $\sum F=-D_{top}+f_s$ will reduce and gradually become closer to zero. The air resistances in this way gradually balances out the equations again and the accelerations gradually decreases.
5. In turn, these decreasing accelerations cause $f_s$ to be smaller, since a smaller force is needed to induce this smaller acceleration in the top book.
6. Soon, $f_s$ and $D_{top}$ meet and become equal in $\sum F=-D_{top}+f_s$ so the total force on the top book equals zero again. This can only happen when also $\sum F=P-D_{bottom}-f_k-f_s$ has balanced out to zero again, because only then are there no more accelerations of the bottom book, which $f_s$ has to induce into the top book as well. $f_s$ is left to take care of only the air resistance on the top book, although at a new value.