The wavefunction vector $|\Psi (t) \rangle $ is supposed to be a function of time only. When you write $ | \Psi (t) \rangle $ you are not considering the projection of the wavefunction nor on the position neither on the momentum space, but just the state of the system at time $ t $, which is nothing but a postulate of Quantum Mechanics. You will have the wavefunction in coordinate (or momentum or any other observable) once you project your state vector on a basis of the observable you have chosen. For instance, in coordinate space: $$ \langle \mathbf{x} | \Psi (t) \rangle := \Psi (\mathbf{x},t)$$.
which is the probability amplitude of finding my system (here we have just one coordinate, so we suppose we are dealing with a single particle system) at position $ \mathbf{x} $ at time $ t$. If you want to switch from coordinate space to momentum space, i.e. you want to have the following probability amplitude: $$ \langle \mathbf{p} | \Psi (t) \rangle = \tilde{\Psi}(\mathbf{p} ,t) $$
(where we have used $ \tilde{\Psi} $ to mean that is not the same function of $ \mathbf{p}$ as $ \Psi$ was for $ \mathbf{x}$), we can write like this: $$ \tilde{\Psi}(\mathbf{p},t)=\int\,d\mathbf{x} \langle\mathbf{p}|\mathbf{x}\rangle\langle\mathbf{x}|\Psi(t)\rangle$$ for each $t$, having inserted the completeness relation of the space coordinate observable.
Now, knowing that $ \langle \mathbf{p} | \mathbf{x} \rangle = \frac{1}{\sqrt{2\pi \hbar}}\exp(\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{x}),$ you find that that projection of wavefunction in momentum space is the fourier transform of the coordinate-space wave function.
As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation.
Recalling that
$$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$
and putting this expression into the (coordinate representation of the) TDSE, we have
$$i\hbar\frac{\partial}{\partial t}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) + V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$
We can move the partials inside the integrals but we must be careful with the potential. Noting that
$$\Phi(p,t) = \int dx \frac{e^{-i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \Psi(x,t)$$
we have
$$V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)$$
But,
$$ \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)= V(p) * \Phi(p,t)$$
where $*$ denotes convolution. Thus, we can write
\begin{align}
\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left(i\hbar\frac{\partial}{\partial t}\Phi(p,t)\right) &= -\int dp \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)\right)\\&\qquad + \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left( V(p) * \Phi(p,t)\right)
\end{align}
leading to
$$\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \left\{ i\hbar\frac{\partial}{\partial t}\Phi(p,t) = -\frac{\hbar^2}{2m}\left(-\frac{p^2}{\hbar^2}\Phi(p,t) \right) + V(p)*\Phi(p,t)\right\}$$
and so
$$i\hbar\frac{\partial}{\partial t}\Phi(p,t) = \frac{p^2}{2m}\Phi(p,t) + V(p)*\Phi(p,t)$$
Best Answer
Yes, you can use eigenstates of the energy operator (the Hamiltonian) to express any state, but
i) The spectrum of the Hamiltonian can be discrete, so that instead of $|\psi\rangle = \int \psi(E)|E\rangle \, dE$, you have $|\psi\rangle = \sum \psi_E |E\rangle$. In fact, in introductory quantum mechanics when introducing the time independent Schrödinger equation, we use that any state can be decomposed as such.
ii) The spectrum of the Hamiltonian can be degenerate, so there is more than one state with the same energy. For example, for a free particle, the energy is $\mathbf p^2/2m$. It depends only on the magnitude of $\mathbf p$, not the direction. Therefore, a formula like $|\psi\rangle = \int \psi(E) |E\rangle$ isn't general enough. Which state with energy $E$ does $|E\rangle$ refer to? You need something like $|\psi\rangle = \int \psi(E, \hat p) |E, \hat p\rangle$ where $\hat p$ is a unit vector and signifies that the state has momentum in the $\hat p$ direction.
Since $E$ is, in this case, a function only of the magnitude of the momentum, we are in effect using the magnitude and direction of the momentum, so this is basically the same as momentum space. It is changing to spherical coordinates in momentum space followed by the change of variable $p \mapsto E = p^2/2m$. This is convenient for example when doing statistical physics because the Bose-Einstein and Fermi-Dirac distributions are functions of the energy, so it's convenient to use the energy as a variable.
iii) The momentum operator is always the same, so its eigenfunctions are always the same, and they are simple; they are plane waves.
On the other hand, the energy, the Hamiltonian operator, is different for each system -- a hydrogen-like atom, a harmonic oscillator, molecules, a crystal, a superconductor... In the general case, finding the eigenstates of the energy is hard and amounts to completely solving the dynamics of the system. You can see this from that if $$|\psi\rangle = \sum \tilde{c}_n(t) |E_n\rangle $$ and $\hat H|E_n\rangle = E_n |E_n\rangle$, then the Schrödinger equation gives $\tilde c_n(t) = c_n \exp(-iE_nt/\hbar)$ (the $|E_n\rangle$ should be time-independent). In the case of degeneracy, the $|E_n\rangle$ can be chosen orthogonal, so that $c_n = \langle E_n |\psi(t=0)\rangle = \int \psi_{E_n}^* \psi$. Hence you can get the time evolution for any initial state, that is, completely solve the dynamics.
iv) In relativistic theories, energy is the time component of the 4-momentum. Since the split between time and space is observer-dependent, in a relativistic theory, going over to momentum space one must also Fourier transform with respect to time. However, because it holds that $E^2 - p^2 = m^2$, only three components in momentum space are actually independent.
To summarize, yes, you can use eigenstates of the energy to expand the state (accounting for discrete and degenerate spectra), but finding the transition to that eigenbasis is equivalent to solving the Schrödinger equation. If you can do that, go ahead. If you cannot, you can still get somewhere with an energy eigenbasis if you only need general properties of the energy. This can be the case in thermodynamics, e.g. superconductors and Landau's theory of Fermi liquids.