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

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

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

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