First note the following: If $\psi(x,0)$ is any normalizable wave-function in your Hilbert space, then $\psi(x,t) = e^{-iHt} \psi(x,0)$ (I've set $\hbar = 1$) satisfies the time-dependent Schrodinger equation (we call this "time evolution"). This can be seen by using the fact that the stationary states form a basis for your Hilbert space and so $\psi(x,0)$ can be expanded in terms of them. One can then check by substitution (after plugging in the expanded form of $\psi(x,0)$ into $\psi(x,t)$) that the TDSE is satisfied.
Now I haven't checked this explicitly, but to me this seems like the time evolved wave function for the ground state shifted to the right by $a$ (this has the right properties at $a=0$ and $t=0$). Namely, we simply apply the time evolution operator to $\psi(x,0) = \psi_0(x-a)$ where $\psi_0(x)$ is the ground state wave function. To actually verify that $\psi(x,t) = e^{-iHt}\psi(x,0)$ coincides with the wave function you write, one can proceed as follows:
Use $$\psi(x,t) = e^{-iHt}\psi(x,0) = \int dy \,K(x,y,t)\psi(y,0),$$ where $K(x,y,t)$ is the quantum mechanical propagator for the harmonic oscillator for which the explicit expression can be found on this Wikipedia link. Perform the subsequent Gaussian integral. If I'm right you'll end up with exactly the wave function that you want.
In short, finding the most general time-dependent solution boils down to finding the propagator. One can then use that to time evolve any stationary wave function that one wishes.
