Does the trace norm of the difference of two density matrices being one imply these two density matrices can be simultaneously diagonalizable?
Here is one way to prove the fact you are interested in.

Suppose $\rho_0$ and $\rho_1$ are density matrices.  Like every other Hermitian matrix, it is possible to express the difference $\rho_0-\rho_1$ as
$$\rho_0-\rho_1 = P_0-P_1$$ for $P_0$ and $P_1$ being positive semidefinite and having orthogonal images.  (Sometimes this is called a Jordan-Hahn decomposition; it is unique and easily obtained from a spectral decomposition of $\rho_0-\rho_1$.) Note that the fact that $P_0$ and $P_1$ have orthogonal images implies that they are simultaneously diagonalizable, which I interpret is the property you are interested in.

The trace norm of the difference $\rho_0-\rho_1$ (as you define it, with the multiplicative factor 1/2), is given by $$\|\rho_0-\rho_1\|_{\text{tr}} = \frac{1}{2}\operatorname{Tr}(P_0) + \frac{1}{2}\operatorname{Tr}(P_1).$$  Under the assumption that this quantity is 1, we will conclude that $P_0=\rho_0$ and $P_1=\rho_1$, which proves what you want to prove.

To draw this conclusion, note first that $\operatorname{Tr}(P_0)-\operatorname{Tr}(P_1)=0$ and $\operatorname{Tr}(P_0)+\operatorname{Tr}(P_1)=2$, so $\operatorname{Tr}(P_0)=\operatorname{Tr}(P_1)=1$.  Next, take $\Pi_0$ and $\Pi_1$ to be the orthogonal projections onto the images of $P_0$ and $P_1$, respectively.  We have $$\Pi_0 (\rho_0 - \rho_1) = \Pi_0 (P_0 - P_1) = P_0$$ so $$\operatorname{Tr}(\Pi_0 \rho_0) - \operatorname{Tr}(\Pi_0 \rho_1) = 1.$$ Both $\operatorname{Tr}(\Pi_0 \rho_0)$ and $\operatorname{Tr}(\Pi_0 \rho_1)$ must be contained in the interval [0,1], from which we conclude that $\operatorname{Tr}(\Pi_0\rho_0)=1$ and $\operatorname{Tr}(\Pi_0\rho_1) = 0$.  From these equations it is not difficult to conclude $\Pi_0\rho_0=\rho_0$ and $\Pi_0\rho_1=0$, and therefore $P_0=\rho_0$ by the equation above.  A similar argument shows $P_1=\rho_1$.