Suppose a manifold is homeomorphic to $\mathbb R^4$, and you've shown that it is equipped with a flat metric of signature $(n-1,1)$. To what extent can I conclude that my space is Minkowski space?
My intuition tells me that there exists something like a linear change of coordinates so that my metric is literally diag$(-1,1,1,1)$. If this is indeed the case, how do I formalise this? I cannot conclude that my metric is the Minkowski metric up to a change in basis, since its eigenvalues need not be in $\{-1,1\}$.
Essentially I'm after a characterisation of Minkowski space that doesn't require me to say "the metric is diag$(-1,1,1,1)$", because I'm working in a more abstract setting and I'd prefer not to compute anything with coordinates.
I am afraid that your intuition is not correct at all, there is a huge variety of flat Lorentzian metrics on $\mathbb R^4$. The main observation here is that the pullback of a flat metric along a diffeomorphism will be flat, too. On the one hand, you can take any gobal diffeomorphism $f:\mathbb R^4\to\mathbb R^4$ and pull back the Minkowski metric along $f$. The result is a complete flat metric on $\mathbb R^4$ (which is globally isometric to Minkowski, but the isometry is given by $f$, so it is by no means a linear change of coordinates).
On the other hand, you can easily construct examples of incomplete flat Lorentzian metrics on $\mathbb R^4$. For example, use a diffeomorphism from $\mathbb R^4$ to the open unit ball in $\mathbb R^4$ to pull back the Minkowski metric on the ball to $\mathbb R^4$. The resulting metric on $\mathbb R^4$ is flat and incomplete by construction, so it cannot be globally isometric to $\mathbb R^4$.
The main thing that you can do in your situation is taking the exponential map from the tangent space in a point to your manifold (which you can identify with Minkowski space). This will define an isometry from an open neighborhood of zero to a neighborhood of the given point. The isometry can be globalized if the metric you start from is complete.