Yes, the density matrix reconciles all quantum aspects of the probabilities with the classical aspect of the probabilities so that these two "parts" can no longer be separated in any invariant way.
As the OP states in the discussion, the same density matrix may be prepared in numerous ways. One of them may look more "classical" – e.g. the method following the simple diagonalization from equation 1 – and another one may look more quantum, depending on states that are not orthogonal and/or that interfere with each other – like equations 2.
But all predictions may be written in terms of the density matrix. For example, the probability that we will observe the property given by the projection operator $P_B$ is
$$ {\rm Prob}_B = {\rm Tr}(\rho P_B) $$
So whatever procedure produced $P_B$ will always yield the same probabilities for anything.
Unlike other users, I do think that this observation by the OP has a nontrivial content, at least at the philosophical level. In a sense, it implies that the density matrix with its probabilistic interpretation should be interpreted exactly in the same way as the phase space distribution function in statistical physics – and the "quantum portion" of the probabilities inevitably arise out of this generalization because the matrices don't commute with each other.
Another way to phrase the same interpretation: In classical physics, everyone agrees that we may have an incomplete knowledge about a physical system and use the phase space probability distribution to quantify that. Now, if we also agree that probabilities of different, mutually excluding states (eigenstates of the density matrix) may be calculated as eigenvalues of the density matrix, and if we assume that there is a smooth formula for probabilities of some properties, then it also follows that even pure states – whose density matrices have eigenvalues $1,0,0,0,\dots$ – must imply probabilistic predictions for most quantities. Except for observables' or matrices' nonzero commutator, the interference-related quantum probabilities are no different and no "weirder" than the classical probabilities related to the incomplete knowledge.
A mixed state is mathematically represented by a bounded, positive trace-class operator with unit trace $\rho : \cal H \to \cal H$.
Here $\cal H$ denotes the complex Hilbert space of the system (it may be nonseparable). The set of mixed states $S(\cal H)$ is a convex body in the complex linear space of trace class operators $B_1(\cal H)$ which is a two-side $*$-ideal of the $C^*$-algebra of bounded operators $B(\cal H)$.
Convex means that if $\rho_1,\rho_2 \in S(\cal H)$ then a convex combination of them, i.e. $p\rho_1 + q\rho_2$ if $p,q\in [0,1]$ with $p+q=1$, satisfies $p\rho_1 + q\rho_2 \in S(\cal H)$.
two-side $*$-ideal means that linear combinations of elements of $B_1(\cal H)$
belong to that space (the set is a subspace), the adjoint of an element of $B_1(\cal H)$ stays in that space as well and $AB, BA \in B_1(\cal H)$ if $A\in B_1(\cal H)$ and $B \in B(\cal H)$.
I stress that, instead, the subset of states $S(\cal H)\subset B_1(\cal H)$ is not a vector space since only convex combinations are allowed therein.
The extremal elements of $S(\cal H)$, namely the elements which cannot be decomposed as a nontrivial convex combinations of other elements, are all of the pure states. They are of the form $|\psi \rangle \langle \psi|$ for some unit vector of $\cal H$. (Notice that, since phases are physically irrelevant the operators $|\psi \rangle \langle \psi|$ biunivocally determine the pure states, i.e. $|\psi\rangle$ up to a phase.)
The space $B_1(\cal H)$ and thus the set $S(\cal H)$ admits at least three relevant normed topologies induced by corresponding norms. One is the standard operator norm $||T||= \sup_{||x||=1}||Tx||$ and the remaining ones are:
$$||T||_1 = || \sqrt{T^*T} ||\qquad \mbox{the trace norm}$$
$$||T||_2 = \sqrt{||T^*T||} \qquad \mbox{the Hilbert-Schmidt norm}\:.$$
It is possible to prove that:
$$||T|| \leq ||T||_2 \leq ||T||_1 \quad \mbox{if $T\in B_1(\cal H)$.}$$
Moreover, it turns out that $B_1(\cal H)$ is a Banach space with respect to $||\cdot||_1$ (it is not closed with respect the other two topologies, in particular, the closure with respect to $||\cdot||$ coincides to the ideal of compact operators $B_\infty(\cal H)$).
$S(\cal H)$ is closed with respect to $||\cdot ||_1$ and, more strongly, it is a complete metric space with respect to the distance $d_1(\rho,\rho'):= ||\rho-\rho'||_1$.
When $dim(\cal H)$ is finite the three topologies coincide (though the norms do not), as a general result on finite dimensional Banach spaces.
Concerning your last question, there are many viewpoints. My opinion is that a density matrix is physical exactly as pure states are. It is disputable whether or not a mixed state encompasses a sort of physical ignorance, since there is no way to distinguish between "classical probability" and "quantum probability" in a quantum mixture as soon as the mixture is created. See my question Classical and quantum probabilities in density matrices and, in particular Luboš Motl's answer.
See also my answer to Why is the application of probability in QM fundamentally different from application of probability in other areas?
ADDENDUM. In finite dimension, barring the trivial case $dim({\cal H})=2$ where the structure of the space of the states is pictured by the Poincaré-Bloch ball as a manifold with boundary, $S(\cal H)$ has a structure which generalizes that of a manifold with boundary. A stratified space. Roughly speaking, it is not a manifold but is the union of (Riemannian) manifolds with different dimension (depending on the range of the operators) and the intersections are not smooth. When the dimension of $\cal H$ is infinite, one should deal with the notion of infinite dimensional manifold and things become much more complicated.
Best Answer
This really depends whether you believe in the "church of the larger Hilbert space". If you feel that pure states are more fundamental than mixed states, then you might argue that any mixed state is just a lack of knowledge, and somewhere out there is the missing piece of the system which will give you full information (i.e., a pure state). Even though you don't know what it is, you know it is out there. (As you can see, this is really more a matter of interpretation of quantum mechanics, since mathematically, the two perspectives are equivalent.)
There are many cases where this is a very reasonable perspective on mixed states regardless of what you believe, e.g. when you have a pure state which becomes mixed by coupling to the environment: While the state of the system looks mixed to you, it has just unitarily interacted with the environment, so the overall state system + environment will be pure, and the environment will hold the purification of your system.
Clearly, this is true more generally as long as your initial global state is pure and you consider part of an isolated sytem (i.e. unitary dynamics).
Beyond that, purifications are of course also a powerful mathematical tool.