Solutions of $\Delta \phi + \phi =0$ on $\mathbb{R}^2$

Every solution of $\Delta\phi+\phi=0$ on $\mathbf R^2$ can be written as the “Poisson integral” (where for short $(u,v)=w$, $(x,y)=z$) $$ \phi(z)= \left\langle T, e^{i\langle z,\cdot\rangle}\right\rangle=\int_{\mathrm S^1}e^{i\langle z,w\rangle}dT(w) $$ for a unique entire functional $\,T$ on the circle $\mathrm S^1$; these include Radon measures, Schwartz distributions, hyperfunctions, and more. Details and proofs see Hashizume et al. (1972, p. 543), Helgason (1974, p. 348; 1984, p. 5), or Agmon (1999). For example, if I am not mistaken,

• $T_1=$ Dirac measure at $(-i,\sqrt2)\in (\mathrm S^1)^{\mathbf C}$ gives $\phi_1(z)=e^{x+\sqrt2iy}$ (found by Maple);
• $T_2=$ derivative of $T_1$ in the direction $(\sqrt2,i)$ gives $\phi_2(z)=(\sqrt2x+iy)e^{x+\sqrt2iy}$ (new).

Regarding the other question in your post: not quite.

If $\phi$ is a bounded function such that $\Delta \phi + \phi = 0$, then $\phi$ is a tempered distribution such that the Fourier transform of $\phi$ (which is another tempered distribution) satisfies $(-|\xi|^2 + 1) \hat\phi = 0$. This easily implies that $\hat\phi$ is supported in the unit sphere $\partial B = \{\xi : |\xi| = 1\}$.

However, $\hat\phi$ need not be a finite measure. If I am not mistaken, the distribution $$ \langle \hat\phi, u \rangle = \lim_{\varepsilon \to 0^+} \int_{\{|\xi_1| > \varepsilon\} \cap \partial B} \frac{u(\xi)}{i \xi_1} \, \sigma(d\xi) $$ (where $\sigma$ is the usual surface measure on $\partial B$) is the Fourier transform a bounded function, the indefinite integral (with respect to $x_1$) of the Bessel function $J_0(|x|)$. (Just as in dimension one the distribution $1/x$ is the Fourier transform of a bounded function.)

The answer to your first question is affirmative. The command of Maple 2019.1

pdsolve(VectorCalculus:-Laplacian(phi(x, y), [x, y]) = -phi(x, y), phi(x, y), explicit);

produces

$$\phi \left( x,y \right) ={\it \_C1}\,{{\rm e}^{\sqrt {{\it \_c}_{{1}}} x}}{\it \_C3}\,\sin \left( \sqrt {{\it \_c}_{{1}}+1}y \right) +{\it \_C1}\,{{\rm e}^{\sqrt {{\it \_c}_{{1}}}x}}{\it \_C4}\,\cos \left( \sqrt {{\it \_c}_{{1}}+1}y \right) +{\frac {{\it \_C2}\,{\it \_C3}\, \sin \left( \sqrt {{\it \_c}_{{1}}+1}y \right) }{{{\rm e}^{\sqrt {{ \it \_c}_{{1}}}x}}}}+{\frac {{\it \_C2}\,{\it \_C4}\,\cos \left( \sqrt {{\it \_c}_{{1}}+1}y \right) }{{{\rm e}^{\sqrt {{\it \_c}_{{1}}} x}}}}. $$