Your equation is the right solution to Schrödinger's equation in the momentum-energy representation. However, it's only that simple for Schrödinger's equation with no potential, $V(x,y,z)=0$.
If it's zero, the solution (or, similarly, the reformulation of the equation) is as easy as the algebraic relationship you wrote - but it's also uninteresting for the same reason. The interesting cases have e.g. the Coulomb potential $k/r$ or the harmonic oscillator potential $kx^2/2$ and they can't be "solved" in the simple way you sketched. For a nonzero potential, the problem is genuinely equivalent to a partial differential equation.
However, that doesn't mean that it's the only way in which the problem may be formulated or solved. Both the harmonic oscillator and the Hydrogen atom may be solved (i.e. their spectrum may be found) algebraically, by the creation and annihilation operators in the harmonic oscillator case, or by a hidden $SO(4)$ symmetry in the Hydrogen atom case.
A general Schrödinger's equation in quantum mechanics is really an ordinary differential equation for the state vector; the "spatial derivatives" only appear as the action of particular operators on the Hilbert space. Some of these operators - e.g. the momenta in the position representations - are conveniently represented as partial derivatives with respect to spatial coordinates but that's only the case if we use a "continuous basis" for the Hilbert space.