This is indeed possible only in some situations, e.g. when the continuous spectrum is absent (it may also consist of a single point, see Valter Moretti's comment below). A sufficient condition for that to be true is that either the Hamiltonian is compact or it has compact resolvent.
Sadly, very few interesting Hamiltonians satisfy that property (an example being the harmonic oscillator). In general, the solution of the Schrödinger equation exists for any initial condition $\psi_0\in\mathscr{H}$ (the Hilbert space), using the unitary one-parameter strongly continuous group $e^{-it H}$ associated to the self-adjoint Hamiltonian $H$ by Stone's theorem. The solution at any time $t\in \mathbb{R}$ is then simply written
$$\psi(t)=e^{-itH}\psi_0\; .$$
Such solution is continuous in time, and with respect to initial data, but it is differentiable in time only if $\psi_0\in D(H)$, where $D(H)$ is the domain of the self-adjoint operator $H$. In the language of analysis of PDEs, that means that the Schrödinger equations, for self-adjoint Hamiltonians, are globally well posed on $\mathscr{H}$.
You are starting from the incorrect point. The argument follows by linearity of the equation.
Suppose $\Psi_k(x,t)$ is solution of the time dependent
Schr$\ddot{\hbox{o}}$dinger equation:
$$
i\hbar \frac{\partial }{\partial t}\Psi_k(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_k(x,t)}{\partial x^2}+U(x)\Psi_k(x,t)\, .
$$
Then:
$$
\Phi(x,t)=a_1\Psi_1(x,t)+a_2\Psi_2(x,t)
$$
is also a solution since
$$
i\hbar \frac{\partial }{\partial t}\Phi(x,t)
=a_1\left(i\hbar \frac{\partial }{\partial t}\Psi_1(x,t)\right)+a_2
\left(i\hbar \frac{\partial }{\partial t}\Psi_2(x,t)\right)
$$
and
\begin{align}
-\frac{\hbar^2}{2m}\frac{\partial^2\Phi(x,t)}{\partial x^2}+U(x)\Phi(x,t)
&=a_1\left(-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_1(x,t)}{\partial x^2}+U(x)\Psi_1(x,t)\right)\\
&\quad + a_2\left(-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_2(x,t)}{\partial x^2}+U(x)\Psi_2(x,t)\right)\, .
\end{align}
These follow simply from the known rule valid for any two differentiable functions $f$ and $g$: $\partial (f+g)/\partial t=\partial f/\partial t+\partial g/\partial t$, and similarly for the partials w/r to $x$.
Combining these last two equations you get an identity for any $a_1$ and $a_2$ since each $\Psi_k(x,t)$ is independently a solution. Of course this simply extends to an arbitrary number of terms in the linear combination.
Note the eigenvalue of the time-independent part never enters in this argument. The final step is to observe that separation of variables in the time-dependent equation yields $\Psi_k(x,t)=e^{-iE_k t}\psi_k(x)$ with $\psi_k(x)$ an eigenfunction of the time-independent equation, but again, this does not enter in the argument.
Edit: note this is in contradistinction with the time-independent equation. When
$$
-\frac{\hbar^2}{2m}\frac{d^2\psi_k(x)}{dx^2}+U(x)\psi_k(x)=E_k\psi_k(x)
$$
the the right hand side is must a multiple of the original function. With this observation, note then
that a linear combination
$$
\psi(x)=a_1\psi_1(x)+a_2\psi_2(x)
$$
will in general NOT be a solution of the time-independent equation because
\begin{align}
\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)\right)\psi(x)
&=a_1\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)\right)\psi_1(x)\\
&\qquad+a_2\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+U(x)\right)\psi_2(x)
\\
&=a_1E_1\psi_1(x)+a_2E_2\psi_2(x)\\
&=E_1(a_1\psi_1(x)+a_2\psi_2(x))+(E_2-E_1)a_2\psi_2(x)\\
&=E_1\psi(x)+(E_2-E_1)a_2\psi_2(x)
\end{align}
will NOT be a multiple of $\psi(x)$ unless $E_1=E_2$.
Best Answer
You have undoubtably heard about Heisenberg's uncertainty principle:
$$ \Delta x \, \Delta p \ge \frac{\hbar}{2} $$
There is a corresponding energy-time uncertainty principle:
$$ \Delta E \, \Delta t \ge \frac{\hbar}{2} $$
The point about the separable wavefunctions is that they are time independant, which means that $\Delta t = \infty$ and therefore that $\Delta E = 0$. This is what Griffiths means by definite energy.
The above argument is rather handwaving, but if you read the part of Griffiths following the excerpt you quote he makes this rigorous. The energy uncertainty, $\Delta E$, is the standard deviation defined by:
$$ (\Delta E)^2 = \langle H^2 \rangle - \langle H \rangle^2 $$
Griffiths shows that this is equal to zero.