When I first faced the issue of in-out asymptotic states reading Bjorken-Drell I've had your same doubt 2) and I've worked out some details trying to disentangle it. Before writing them down, I should mention that other answers specify that product state are not eigenstate of the interacting four-momentum $P^\mu$ (of which $P^0$ is the interacting Hamiltonian) and advocate to more advanced mathematical treatment of scattering theory in the context of axiomatic QFT. Instead my answer only refers to (and thus is valid only at the level of rigor of) the treatment of asymptotic states/evolution given in Bjorken-Drell type/level books (and only the case of a single scalar field involved), even though I think that previous comments should have specified that the (non-normalizable) asymptotic states are not (generalized) eigenstate of the S-matrix (which describes the evolution for an infinite amount of time), but they are in fact eigenstates of the interacting four-momentum $P^\mu$, as it clearly is stated in Bjorken-Drell section 16.3, formulas 16.10 and 16.11 (valid for product and non-product state), i.e

$[P^\mu, a_{in}(k)]=-k^\mu a_{in}(k)\;$ and $\;[P^\mu, a^\dagger_{in}(k)]=k^\mu a^\dagger_{in}(k)\;$ (16.10)

$P^\mu |k_1,...,k_n,in>=(\sum_{i=1}^n k_i) |k_1,...,k_n,in>\;$ (16.11)

where $|k_1,...,k_n, in>= a^\dagger_{in}(k_1)...a^\dagger_{in}(k_n)|0>$, $\;a^\dagger_{in}(k)\;$ and $\;a_{in}(k)$ are creation/destruction operators for in single particle states with four momentum $k$ and $|0>$ is the interacting vacuum. The same is valid for out asymptotic states $\;|k_1,...,k_m, out>= a^\dagger_{out}(k_1)...a^\dagger_{out}(k_m)|0>$, i.e they are eigenstate of the four-momentum $P^\mu$ but they are not the same set of eigenstate as the in-eigenstates $|k_1,...,k_n, in>$ (unless $m=n=1$, i.e $|k, in>=|k, out>$ only for one particle states).

This is the crucial point to understand and the help comes from section 16.6 of Bjorken-Drell's book. In this section the authors introduced the formalism of S-matrix operator, specifying definition and basic properties. The definition of $S$ implies that one can map in-eigenstates $|\alpha, in>=|k_1,...,k_n, in>$ to out-eigenstates $|\beta,out>=|k_1,...,k_m, out>$ through the $S$ operator, i.e $|\beta,out>=S^\dagger|\alpha, in>$ and the other significant property is that $S$ is unitary: this means that $S$ is nothing more than a unitary change of basis between the complete orthonormal set of in-eigenstates $|\alpha, in>$ and the complete orthonormal set of out-eigenstates $|\beta,out>$. The main question mark with this interpretation is that in fact the two complete orthonormal set of in/out-states must be eigenstates of the same 4-momentum operator $P^\mu$: the only possible eventuality when we can have two different set of eigenstates for the same operator is if this operator is degenerate (at least one of it's eigenspaces have dimension greater than one). Then the question is: Is $P^\mu$ degenerate? The answer is yes and it's easy to see it: if we apply $P^\mu$ to $|k_1,...,k_n, in>$, we get $P^\mu |k_1,...,k_n,in>=(\sum_{i=1}^n k_i) |k_1,...,k_n,in>$ and the same eigenvalue $\sum_{i=1}^n k_i$ of $P^\mu$ is associated with every possible state $|k'_1,...,{k'}_{n'},in>$ which satisfy $\sum_{i=1}^{n'} k'_i=\sum_{i=1}^n k_i$ (thus total momentum conservation is automatically implied in the fact that, through the S-matrix, we can only transform the basis in the degenerate eigenspace of fixed total momentum to make it possible that in and out state are eigenstates of the same operator $P^\mu$). I hope this clarify the issue.

Unfortunately I can't directly interact with the authors of those answers I mentioned in the beginning of this post as I can't leave comments yet.

When I first faced the issue of in-out asymptotic states reading Bjorken-Drell I've had your same doubt 2) and I've worked out some details trying to disentangle it. Before writing them down, I should mention that other answers specify that product state are not eigenstate of the complete (*) four-momentum $P^\mu$ (of which $P^0$ is the interacting Hamiltonian) and advocate to more advanced mathematical treatment of scattering theory in the context of axiomatic QFT. Instead my answer only refers to (and thus is valid only at the level of rigor of) the treatment of asymptotic states/evolution given in Bjorken-Drell type/level books (and only the case of a single scalar field involved). The (non-normalizable) asymptotic states are in fact (generalized) eigenstates of the complete four-momentum $P^\mu$, as it clearly is stated in Bjorken-Drell section 16.3, formulas 16.10 and 16.11, i.e

$[P^\mu, a_{in}(k)]=-k^\mu a_{in}(k)\;$ and $\;[P^\mu, a^\dagger_{in}(k)]=k^\mu a^\dagger_{in}(k)\;$ (16.10)

$P^\mu |k_1,...,k_n,in>=(\sum_{i=1}^n k_i) |k_1,...,k_n,in>\;$ (16.11)

where $|k_1,...,k_n, in>= a^\dagger_{in}(k_1)...a^\dagger_{in}(k_n)|0>$, $\;a^\dagger_{in}(k)\;$ and $\;a_{in}(k)$ are creation/destruction operators for in single particle states with four momentum $k$ and $|0>$ is the vacuum of the complete hamiltonian $H=H_0+H_i$. The same is valid for out asymptotic states $\;|k_1,...,k_m, out>= a^\dagger_{out}(k_1)...a^\dagger_{out}(k_m)|0>$, i.e they are eigenstate of the four-momentum $P^\mu$ but they are not the same set of eigenstate as the in-eigenstates $|k_1,...,k_n, in>$ (unless $m=n=1$, i.e $|k, in>=|k, out>$ only for one particle states).

This is the crucial point to understand and the help comes from section 16.6 of Bjorken-Drell's book. In this section the authors introduced the formalism of S-matrix operator, specifying definition and basic properties. The definition of $S$ implies that one can map in-eigenstates $|\alpha, in>=|k_1,...,k_n, in>$ to out-eigenstates $|\beta,out>=|k_1,...,k_m, out>$ through the $S$ operator, i.e $|\beta,out>=S^\dagger|\alpha, in>$ and the other significant property is that $S$ is unitary: this means that $S$ is nothing more than a unitary change of basis between the complete orthonormal set of in-eigenstates $|\alpha, in>$ and the complete orthonormal set of out-eigenstates $|\beta,out>$. The main question mark with this interpretation is that in fact the two complete orthonormal set of in/out-states must be eigenstates of the same 4-momentum operator $P^\mu$: the only possible eventuality when we can have two different set of eigenstates for the same operator is if this operator is degenerate (at least one of it's eigenspaces have dimension greater than one). Then the question is: Is $P^\mu$ degenerate? The answer is yes and it's easy to see it: if we apply $P^\mu$ to $|k_1,...,k_n, in>$, we get $P^\mu |k_1,...,k_n,in>=(\sum_{i=1}^n k_i) |k_1,...,k_n,in>$ and the same eigenvalue $\sum_{i=1}^n k_i$ of $P^\mu$ is associated with every possible state $|k'_1,...,{k'}_{n'},in>$ which satisfy $\sum_{i=1}^{n'} k'_i=\sum_{i=1}^n k_i$ (thus total momentum conservation is automatically implied in the fact that, through the S-matrix, we can only transform the basis in the degenerate eigenspace of fixed total momentum to make it possible that in and out state are eigenstates of the same operator $P^\mu$). I hope this clarify the issue.

Edit (*): I wrote "interacting four momentum" but must have written "complete four-momentum", because this can generate confusion with the interacting picture 4-momentum, instead the momentum used by B&D in chapter 16 is the complete 4-momentum $P^\mu=P^\mu_0+P^\mu_i$ of the theory (free+interaction part).