I'm changing the names of the coordinates slightly to $(t,x)$. The light cone $K:=\{(t,x)\in\mathbb{R}^{1+n} \mid t\geq 0 \wedge t^2 - \|x\|^2 \geq 0 \}=\{(t,x) \mid t\geq \|x\|\}$ is a closed, convex, cone. Consider the dual cone $K^\vee := \{(s,y) \mid st+\langle y,x\rangle \leq 0\}=\{(s,y) \mid s\leq 0 \wedge s^2-\|y\| \geq 0\}$.

A general form of the Paley-Wiener theorem tells you that any for tempered distribution $u\in\mathcal{S}'(\mathbb{R}^{1+n})$ with $supp(u)\subseteq K$, the Fourier transform $\mathcal{F}u$ extends to a holomorphic function $F: \mathbb{R}^n + i (K^\vee)^\circ \to \mathbb{C}$ that satisfies a certain growth condition. Here "extends" means that if you consider the "slice" functions $F_b:\mathbb{R}^n\to\mathbb{C}, a\mapsto F(a+ib)$, then $\lim_{t\searrow 0} F_{tb} \to \mathcal{F}u$ for $b$ in the interior of $K^\vee$.

Note that $F_b=\mathcal{F}(e^{\langle b,x\rangle}u(x))$.

This is not quite the situation of the identity theorem or the edge-of-the-wedge theorem, but it's close enough that I think one can show from this that $F$ must be constant everywhere and therefore zero so that $u=0$.

I'm changing the names of the coordinates slightly to $(t,x)$. The light cone $K:=\{(t,x)\in\mathbb{R}^{1+n} \mid t\geq 0 \wedge t^2 - \|x\|^2 \geq 0 \}=\{(t,x) \mid t\geq \|x\|\}$ is a closed, convex, cone. Consider the dual cone $K^\vee := \{(s,y) \mid st+\langle y,x\rangle \leq 0\}=\{(s,y) \mid s\leq 0 \wedge s^2-\|y\| \geq 0\}$.

A general form of the Paley-Wiener theorem tells you that any for tempered distribution $u\in\mathcal{S}'(\mathbb{R}^{1+n})$ with $supp(u)\subseteq K$, the Fourier transform $\mathcal{F}u$ extends to a holomorphic function $F: \mathbb{R}^n + i (K^\vee)^\circ \to \mathbb{C}$ that satisfies a certain growth condition. Here "extends" means that if you consider the "slice" functions $F_b:\mathbb{R}^n\to\mathbb{C}, a\mapsto F(a+ib)$, then $\lim_{t\searrow 0} F_{tb} =\mathcal{F}u$ (limit w.r.t. the strong topology on $\mathcal{S}'$) for $b$ in the interior of $K^\vee$.

Note that $F_b=\mathcal{F}(e^{\langle b,x\rangle}u(x))$.

This is not quite the situation of the identity theorem or the edge-of-the-wedge theorem, but it's close enough that I think one can show from this that $F$ must be constant everywhere and therefore zero so that $u=0$.