Can you show that: If for some $1\leq p\leq \infty$ function $f\in L^p(\mathbb{R}^n)$ solves $\Delta f-\lambda^2 f=0$ then $f\equiv 0$. (This is essentially uniqueness of solution to homogenous Yukawa's equation but proof isnt clear to me)
$\lambda>0$
I think this works: Consider $f$ a tempered distribution, which solves $$ -\Delta f +\lambda^2 f =0, $$ in the sense of (tempered) distributions. Taking the Fourier transform gives that $$ \left(| \xi |^2 + \lambda^2 \right) \hat{f} =0. $$ Since a smooth function $\varphi$ is in the Schwartz space $\mathcal{S}$ if and only if $(| \xi|^2 + \lambda^2 ) \varphi \in \mathcal{S}$, the last equation implies that $\hat{f}=0$. Since Fourier transform is an isomorphism we get $f=0$.
Since $L^p(\mathbb{R}^n) \subset \mathcal{S}'$ for every $1\leq p\leq \infty$ this proves what you want.