Invariance of $ds^2$ from invariance of all null intervals
Proposition. In a space $V\cong \mathbb{R}^n$ of dimension $n=p+q$ let there be given an indefinite metric tensor $$\eta~=~\begin{pmatrix} \mathbf{1}_{p\times p}& \mathbf{0}_{p\times q} \cr \mathbf{0}_{q\times p} & -\mathbf{1}_{q\times q} \end{pmatrix}_{n\times n}~=~{\rm diag}(\underbrace{+1,\ldots,+1}_{p\text{ times}},\underbrace{-1,\ldots,-1}_{q\text{ times}})\tag{1}$$ of signature $(p,q)$, and a (possibly degenerate & indefinite) metric tensor $g$. Assume that all null-vectors for $\eta$ are also null-vectors for $g$: $$\forall v\in V :~~v^t\eta v~=~0~~\Rightarrow~~v^tg v~=~0.\tag{2}$$ Then $g$ is proportional to $\eta$: $$\exists \lambda\in \mathbb{R}:~~g~=~\lambda \eta.\tag{3}$$
Sketched proof of proposition:
Let $e_i=(0,\ldots, 0,1,0,\ldots, 0)^t$ be the $i$th unit-vector (of $\eta$-length-square $\pm 1$). Write the metric tensor $$g~=~ \begin{pmatrix} a& b^t \cr b & c \end{pmatrix} \tag{4}$$ in terms of a symmetric $p\times p$ matrix $a$, a symmetric $q\times q$ matrix $c$, and a rectangular $q\times p$ matrix $b$.
Use the following "polarization trick" to show that the $b$-block vanishes: $$b~=~0.\tag{5}$$ If $g_{ij}=e_i^tge_j$ corresponds to a matrix element in the $b$-block, then $v_{\pm}:=e_i\pm e_j$ are null-vectors, so $$4g_{ij}~=~4e_i^tge_j~=~(v_++v_-)^tg(v_+-v_-)~=~v_+^tgv_+ -v_-^tgv_-~\stackrel{(2)}{=}~0+0~=~0. \tag{6}$$
We can diagonalize the symmetric $a$ and $c$ blocks by orthogonal matrices while keeping $\eta$ invariant. In other words, we may assume w.l.o.g. that $$g\text{ is diagonal}.\tag{7}$$
Finally, by considering null-vectors of the form $v:=e_i+ e_j$, it becomes clear that $$g_{ii}+g_{jj}~=~e_i^tge_i+e_j^tge_j~\stackrel{(7)}{=}~v^tgv~\stackrel{(2)}{=}~0.\tag{8}$$ This implies that both $a$ and $c$ are proportional to an identity matrix. The sought-for eq. (3) follows. $\Box$