The issue of the Moon's tidal gravitational effects actually came up some months earlier in connection to the 'leap second' added to compensate for a slowing Earth's rotation, e.g. http://brane-space.blogspot.com/2012/07/where-did-that-extra-second-come-from.html
Still, the Earth's period for one rotation, 86,400s (or 86, 401s on the day the leap second was added) is much much shorter than the Moon's synodic period (29.5 days) or sidereal period (27.3 days). This leads to a situation such as depicted in the attached diagram. Here, I denote two tidal bulges on the Earth's surface by x and x' at essentially opposite ends of the line through the center of Earth. Many blog readers probably already know that two tidal bulges occur as a result of a differential gravitational force. (The general origin for a given tidal force on Earth due to the Moon is from the variable values for the Moon's gravitational attraction at differing locations inside Earth.)
In the sketch I've shown, note that bulge x is closer to the Moon than bulge x' on the other side of Earth. This difference results in a net torque exerted on Earth (torque is the product of the moment of inertia, I, and the angular acceleration, a). Specifically for our diagram, bulge x leads the Moon, and because it's also nearer the Moon than bulge x' , the force exerted by the Moon on x is greater yielding a net torque. This net torque slows the Earth's rotation, in effect, slows its angular momentum. Now, at the same time, bulge x is pulling the Moon forward in its orbital path, effectively speeding it up and causing it to move farther out. This is just a consequence of the conservation of angular momentum.
To fix this concept, consider two bodies m and M connected as shown with c the center of mass:
Mass M O---r1------c ---------r2----------o m
Both M and m rotate about c at distances r1 and r2 respectively. Since the angular momentum, e.g. L = Iw must be constant (where w = v/r)then if m somehow speeds up (higher v) then to compensate, r2 must increase. (w ~ 1/r and w ~ v) . Thus, the Moon constantly moves further out, and by about 3-4 cm a year.
Now, consider Phobos - which orbital period is shorter than a Martian day,since tidal deceleration is decreasing its orbital radius at the rate of about 20 metres (66 ft). This is the converse of the case for our Moon. Thus, Phobos (orbiting faster then Mars is rotating) experiences the opposite of what our Moon does, i.e. a slowing down of its orbital speed. Again, by the law of conservation of angular momentum (and using the simple model shown above), if v is decreased then r must decrease. Mars' other moon Deimos, meanwhile, moves in its orbit faster than Mars is rotating and therefore speeds up like our own Moon, and hence increases in its orbital distance.
Therefore, the conjecture that if our own Moon "just happened to form" at the geosynchronous distance the orbit would remain fixed and stable, is technically correct provided position only is considered, irrespective of dynamical interactions.. Since its "positionally assigned" orbital velocity is in synch with Earth's rotational velocity there is no net torque acting on Earth and hence, the Moon could still be orbiting. The downside is that such a fortuitous happenstance has about as much probability as happening as a blind man playing eight -ball pool and getting all of his balls in before his opponent can!
Hence, a definitive answer to the question would hinge on: 1) the exact mass of the object captured, 2) the Lagrange point solutions based on this mass, and most importantly, 3) the assumptions used in considering the nature of the problem, i.e. two-body (ignoring Mars' small moons Deimos and Phobos entirely), or restricted 3-body, i.e. considering only the more massive, say Phobos.
Re: "the Lagrangian points", in general there are 5 such points in all which refer to the five positions for a hypothetical orbit for which a small object (i.e. tiny moon) can be incorporated without altering the overall Lagrangian pattern (which defines 5 different points at high and low gravitational potential, i.e. V = -GM/r).
Readers who wish to learn more can google "Lagrangian points".