Could this problem be treated so that one body is standing still and the other is orbiting it in an elliptical orbit?
Yes. That's the point of using reduced mass.
If so, where is the focus of this ellipse?
It's an ellipse with one of the foci at the body that is standing still.
There are a number of pathways to get to the same result. An easy one is via Newton's laws of motion. Without loss of generality, I'll work in the center of mass frame of the two objects (which presumably is an inertial frame). Denoting the particles as particle 1 and particle 2, the magnitude of the force that each exerts on the other is
$$F=\frac{Gm_1m_2}{r^2}$$
where $r$ is the magnitude of the distance between the two particles. Denoting $\hat r$ as the unit vector in the direction of particle 2 from the center of mass, the accelerations of the two particles are
$$\begin{aligned}
\frac{d^2}{dt^2} \vec r_2 &= -\frac{Gm_1}{r^2} \hat r \\
\frac{d^2}{dt^2} \vec r_1 &= \phantom{-}\frac{Gm_2}{r^2} \hat r \\
\end{aligned}$$
This is a bit pesky. The right hand sides are functions of $r$ rather than $\vec r_2$ and $\vec r_1$. Taking the difference between the yields
$$\frac{d^2}{dt^2} (\vec r_2 - \vec r_1) = -\frac {G(m_1+m_2)}{r^2} \hat r$$
Denoting $\vec r = \vec r_2 - \vec r_1$, this yields
$$\frac{d^2}{dt^2} \vec r = -\frac {G(m_1+m_2)}{r^2} \hat r = -\frac {G(m_1+m_2)}{||\vec r||^3} \vec r$$
Now we have a nice simple system. The $r$ and $\hat r$ on the right hand side are functions of the $\vec r$ on the left, specifically, $r = ||\vec r||$ and $\hat r = \vec r/r$. This vector $\vec r$ is the vector pointing from particle 1 to particle 2. In other words, the above describes the motion of particle 2 with respect to particle 1.
Note that this is a second order differential equation of the form
$$\frac{d^2 \vec r(t)}{dt^2} = -\frac k {r^2} \hat r = -\frac k {||\vec r||^3} \vec r$$
The only solutions to such ODEs are conic sections: a circle, an ellipse, a parabola, or a hyperbola. Thus the trajectory of one body with respect to another is a conic section.
How about the trajectory of a single body with respect to the center of mass? Using $\vec r = \left(1+\frac {m_2}{m_1}\right)\vec r_2$, the differential equation becomes
$$\frac{d^2}{dt^2} \vec r_2 = -\frac {G(m_1+m_2)}{\left(1+\frac {m_2}{m_1}\right)^3 ||\vec r_2||^3} \vec r_2$$
Note that this is still of the form $\ddot {\vec x} = -k\vec x/||\vec x||^3$. The motion of an individual body with respect to the center of mass is also a conic section.
Lagrangian formulation
The concept of reduced mass is motivated by looking at the kinetic energy of the system. Once again working in the center of mass frame, the kinetic energy of the system is $T = \frac12 m_1 \dot {\vec {r_1}}^2 + \frac12 m_2 \dot {\vec {r_2}}^2$. Expressing $\dot {\vec {r_1}}$ and $\dot {\vec {r_2}}$ in terms of $\dot{\vec r}$ yields
$$\begin{aligned}
\dot {\vec {r_2}} &= \phantom{-}\frac{m_1}{m_1+m_2} \dot{\vec r} \\
\dot {\vec {r_1}} &= -\frac{m_2}{m_1+m_2} \dot{\vec r} \\
T &= \frac12 \frac {m_1m_2}{m_1+m_2} \dot {\vec r}^2
\end{aligned}$$
Defining the reduced mass $m_r = \frac {m_1m_2}{m_1+m_2}$ yields $T = \frac 12 m_r \dot {\vec r}^2$. This is the kinetic energy of a single particle of mass $m_r$ moving at velocity $\dot {\vec r}$. The Lagrangian for a pair of interacting particles in a central force field is thus
$L = \frac 1 2 m_r v^2 - U(r)$
This once again reduces the two body problem to that of a single body. It is in studies of the above with $U(r) = -k/r$ (inverse square law) that the conic section nature of the solutions are uncovered. That takes multiple pages; I am not reproducing it here. You can find these derivations in practically any classical mechanics physics text. The books by Marion and by Taylor are reasonably good undergraduate texts. Goldstein is an introductory graduate level text.
