Let $\mathscr{A}$ be an abelian category. Then, taking an arrow category does not commute with taking derived categories and this is the main point of a derivator - taking an arrow category first allows for better homotopical properties.
So in general, $D(\mathscr{A}^{[1]}) \not \simeq D(\mathscr{A})^{[1]}$. But why is this true?
I'm not sure what kind of invariants to consider (if any) to distinguish these two categories.
On
It's not so interesting to ask for these categories to be inequivalent, though they surely are. The point is rather than the obvious functor is not an equivalence, which means we can't use constructions of cones and such on $D(\mathcal A^{[1]})$ to get such constructions on $D(\mathcal A)^{[1]}.$ The obvious functor is not an equivalence because it is not faithful. Intuitively, this is since morphisms in $D(\mathcal A^{[1]})$ are only equal if there is a homotopy between their defining squares, while morphisms in $D(\mathcal A^{[1]})$ only require homotopies along the edges, which need not cohere together.
Amazingly, according to MO/379107 the arrow category $\mathscr{T}^{[1]}$ of a triangulated category $\mathscr{T}$ has no triangulated structure for $\mathscr{T} \neq 0$.
But $D(\mathscr{A}^{[1]})$ admits a triangulated structure and by the above comment $D(\mathscr{A})^{[1]}$ in general is not. So they cannot be equivalent as categories.