Short answer: yes, they're equally real, and they combine to form the full wavefunction for the system.
The full wavefunction $\Psi(\mathbf{r}_e, \mathbf{r}_p)$ for an electron and proton interacting with each other is the solution to the 6-variable PDE
$$
- \frac{\hbar^2}{2 m_e} \nabla^2_{r_e} \Psi - \frac{\hbar^2}{2 m_p} \nabla^2_{r_e} \Psi - \frac{k e^2}{|\mathbf{r}_e - \mathbf{r}_p|}\Psi = E \Psi,
$$
where $m_e$ and $m_p$ are the respective masses of the electron and proton, $\mathbf{r}_e$ and $\mathbf{r}_p$ are their respective positions, and $\nabla^2_{r_e}$ stands for the Laplacian involving derivatives with respect to $\mathbf{r}_e$ (and similarly for $\nabla^2_{r_p}$.)
Trying to solve this equation as it stands is pretty much impossible. There are six different variables in play (three for each particle) and all of the equations are coupled together via the potential term. However, a clever change of coordinates can make our lives simpler. We can define
\begin{align*}
\mathbf{R} &= \frac{m_e \mathbf{r}_e + m_p \mathbf{r}_p}{M} = \text{position of the center of mass} \\
\mathbf{r} &= \mathbf{r}_e - \mathbf{r}_p = \text{separation vector between electron & proton}
\end{align*}
where $M = m_e + m_p$. If we rewrite the Schrödinger equation in terms of these variables, it becomes
$$
- \frac{\hbar^2}{2 M} \nabla^2_{R} \Psi - \frac{\hbar^2}{2 \mu} \nabla^2_{r} \Psi - \frac{k e^2}{r}\Psi = E \Psi,
$$
where $\mu$ is the reduced mass defined via $\mu^{-1} = m_e^{-1} + m_p^{-1}$.
Why does this help us? The PDE is separable now! Specifically, we can look for a solution of the form
$$
\Psi(\mathbf{r}_e, \mathbf{r}_p) = \psi(\mathbf{R}) \phi(\mathbf{r})
$$
where the functions $\psi$ and $\phi$ satisfy
$$
- \frac{\hbar^2}{2 M} \nabla^2_{R} \psi = E_R \psi \\
- \frac{\hbar^2}{2 \mu} \nabla^2_{r} \phi - \frac{k e^2}{r}\phi = E_r \phi
$$
If we can find these solutions, then their product will satisfy the full 6-D Schrödinger equation with $E = E_R + E_r$. But the solutions to the first equation are just plane waves in $\mathbf{R}$, while the solutions to the second equation are the familiar solutions for the hydrogen atom. So the energy eigenstates of the system would be of the form
$$
\Psi(\mathbf{r}_e, \mathbf{r}_p) = e^{i \mathbf{k} \cdot \mathbf{R}} R(r) Y(\theta,\phi)
$$
with $\mathbf{R}$ and $\mathbf{r}$ defined in terms of $\mathbf{r}_e$ and $\mathbf{r}_p$ as above.
You could then construct a Gaussian wavepacket by taking a superposition of several such plane waves in $\mathbf{R}$. The resulting wavefunction could then be used to predict (say) the expected position of the electron, by calculating the integral
$$
\langle \mathbf{r}_e \rangle = \int \mathbf{r}_e \Psi^*(\mathbf{r}_e, \mathbf{r}_p) \Psi(\mathbf{r}_e, \mathbf{r}_p) \, d^3 \mathbf{r}_e \,d^3 \mathbf{r}_p
$$
Note that both the "plane-wave" and "hydrogenic" parts of $\Psi$ would need to be taken into account, since they both depend on $\mathbf{r}_e$ via their dependence on $\mathbf{R}$ or $\mathbf{r}$.
