The MazurUlam theorem says that any surjective isometry of normed vector spaces is affine. This argument doesn't seem to apply to Minkowski space (of special relativity) since the metric is degenerate. How would one show that the Poincaré group consists of affine maps? This seems really standard but I can't seem to find it anywhere.

Let's fix notation and define the bilinear form $\eta: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R}$ by: $\eta((x,y,z,t),(x',y',z',t')) = xx'+yy'+zz' tt'$ Given a map $T:\mathbb{R}^4 \to \mathbb{R}^4$ which fixes $0$ and preserves $\nu$ we want to show that $T$ is linear. Let $e_1,e_2,e_3,e_4$ be the canonical basis of $\mathbb{R}^4$. The first observation is that for any four vectors $v_1,v_2,v_3,v_4$ such that $\eta(v_i,v_j) = \eta(e_i,e_j)$ for all $i,j$ the linear map sending each $v_i$ to $e_i$ is invertible and preserves $\nu$. Hence by composing $T$ with a linear invertible $\nu$ preserving map we may assume that $Te_i = e_i$ for $i = 1,2,3,4$. Now we have for any $v \in \mathbb{R}^4$ that $\eta(v,e_i) = \eta(Tv,Te_i) = \eta(Tv, e_i)$ this implies that $T$ is the identity (since we can get each coordinate of $Tv$). 


The following paper shows that if chronological order on $\mathbb R^n$ is defined by cone (i.e., $x\in \mathbb R^n$ chronologically precedes $y\in \mathbb R^n$ iff $y − x$ belongs to some fixed cone) then any bijection which preserve the chronological order has to be linear. This statement is much stronger than you need. After Alexandrov, it was reproved independently 5 times or so. 


I haven't worked out the details, so I might have this all wrong, but couldn't you proceed as follows:
It seems to me that the proof that an isometry is linear should be very similar to the proof that any flat metric is locally isometric to the standard one. 

