There are a few things that keep Saturn's rings roughly the way they are.
First, Saturn's D ring actually is "raining" down on Saturn currently. But, the phenomenon of shepherd moons prevents the vast majority of material from leaving the other rings: "The gravity of shepherd moons serves to maintain a sharply defined edge to the ring; material that drifts closer to the shepherd moon's orbit is either deflected back into the body of the ring, ejected from the system, or accreted onto the moon itself." (quote from Wikipedia)
Besides this, the majority of the particles within the ring system have almost no motion towards or away from Saturn; no motion towards the planet prevents them from being lost.
Second, Saturn's rings cannot clump into "full-fledged" moons, but they can clump into moonlets up to several hundred meters to a few kilometers across. At last count, I think there were over 200 that had been found, and they also come out of numerical simulations.
Beyond these larger moonlets, quasi-stable clumps and clusters of ring particles form with great frequency the farther you get from Saturn. These clusters of particles are constantly changing size, trading material, etc., and so there's no time for them to become solid and cohesive.
This gets into the idea of the Roche Limit and Hill Spheres. The basic idea of the Roche Limit is that the closer you are to a massive object, the more tidal forces are going to tear you apart (or prevent you from forming to begin with). Hill spheres are related, where the idea is at what point you're gravitationally bound to one object or another. If you're within Saturn's Hill sphere versus a moon's Hill sphere, you're going to be pulled to Saturn. With both concepts, you'll need to have a moon forming farther away from Saturn than its rings are now to actually be stable.
You can see the effects of these by looking at N-body dynamical simulations of the rings. This was my research for a year and a half, and it culminated in over a hundred simulations, many of which I made movies of, and then I posted them on one of my personal websites. If you go to it, scroll down and take a look at one of the C ring simulations, B ring simulations, and A ring simulations (warning - the movies are a bit big). You should choose ones with a large τ value and ρ of 0.85 because those will show clumping better.
What you'll see is that, in the C ring, almost no clumping occurs. Go farther from Saturn into the B ring and you'll see a spider web start to happen of strands of clumps of particles. Then if you go to the farther away A ring the strands are fragmented more into clusters. (Note on the movies: The "L" value next to each one is how large the simulation cell is on a side, in meters. So you're just looking at a VERY small region of the ring. It's set so that the center of the cell doesn't move, so you'd imagine that whole thing orbiting around Saturn.)
The Kepler orbit of the Earth around the Sun is determined by two constants: the
specific orbital energy $E$ and the specific relative angular momentum $h$:
$$
\begin{align}
E &= \frac{1}{2}v_{r,\oplus}^2 + \frac{1}{2}v_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\
h^2 &= r^2\,v^2_{T,\oplus} = \mu a(1-e^2),
\end{align}
$$
where $\mu = G(M_\odot + M_\oplus)$, $r$ is the distance Earth-Sun (at the moment of impact), $a$ is the semi-major axis, $e$ is the orbital eccentricity, $v_{r,\oplus}$ is the radial orbital velocity of the Earth, and $v_{T,\oplus}$ the tangential velocity. Now, suppose that a large asteroid collides with the Earth, with orbital velocity $(v_{T,A},v_{r,A})$ and mass $M_A$. Its relative velocity is then
$$
\begin{align}
v_{T,A}' &= v_{T,A} - v_{T,\oplus},\\
v_{r,A}' &= v_{r,A} - v_{r,\oplus}.
\end{align}
$$
We can express these relative velocities in terms of the total impact velocity $v_\text{i}$ and the impact angle $\theta$:
$$
\begin{align}
v_{T,A}' &= v_\text{i}\cos\theta,\\
v_{r,A}' &= -v_\text{i}\sin\theta,
\end{align}
$$
where I defined $\theta$ as in Fig. 1 of this article. So we obtain
$$
\begin{align}
v_{T,A} &= v_{T,\oplus} + v_\text{i}\cos\theta,\\
v_{r,A} &= v_{r,\oplus} - v_\text{i}\sin\theta.
\end{align}
$$
If we assume that the collision is central, that heat loss is negligible and that the debris remains gravitationally bound to the Earth, then conservation of momentum implies
$$
\begin{align}
M_\oplus\,v_{T,\oplus} + M_A\,v_{T,A} &= (M_\oplus+M_A)u_{T,\oplus}\\
M_\oplus\,v_{r,\oplus} + M_A\,v_{r,A} &= (M_\oplus+M_A)u_{r,\oplus},
\end{align}
$$
with $(u_{T,\oplus},u_{r,\oplus})$ the new orbital velocity of the Earth (and the gravitationally bound debris) after the impact. We get
$$
\begin{align}
u_{T,\oplus} &= v_{T,\oplus} + \frac{M_A}{M_\oplus+M_A}v_\text{i}\cos\theta,\\
u_{r,\oplus} &= v_{r,\oplus} - \frac{M_A}{M_\oplus+M_A}v_\text{i}\sin\theta.
\end{align}
$$
So the orbital energy and angular momentum will have changed into
$$
\begin{align}
E' &= \frac{1}{2}u_{r,\oplus}^2 + \frac{1}{2}u_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a'},\\
h'^2 &= r^2\,u^2_{T,\oplus} = \mu a'(1-e'^2).
\end{align}
$$
(the change in $\mu$ is negligible). Right, let's plug in some numbers. Suppose we start with a circular orbit, with a radius equal to the present-day semi-major axis:
$$
\begin{align}
\mu &= 1.32712838\times 10^{11}\;\text{km}^3\,\text{s}^{-2},\\
r &= a = 1.49598261\times 10^{8}\;\text{km},\\
e &= 0.
\end{align}
$$
For a circular orbit, it follows that
$$
\begin{align}
v_{T,\oplus} &= \sqrt{\frac{\mu}{r}}= 29.785\;\text{km}\,\text{s}^{-1},\\
v_{r,\oplus} &=0\;\text{km}\,\text{s}^{-1}.
\end{align}
$$
The impact velocity of an asteroid will always be at least equal to the Earth's escape velocity $11.2\,\text{km/s}$, which is the speed it takes up as it falls into the Earth's gravitational potential well. The article that I already linked to states that typical asteroid impact velocities are in the range of $12-20\,\text{km/s}$. In theory, the impact velocity can be as large as $72\,\text{km/s}$ in the case of a head-on collision, when the Earth and the asteroid have opposite orbital velocities, thus a relative velocity of ~$60\,\text{km/s}$, augmented with the escape velocity as the asteroid falls into our gravitational potential well. This is very unlikely for asteroids, but it is possible for comets.
So, let us assume a typical impact velocity $v_\text{i}=16\,\text{km/s}$,
a mass $M_A = 0.1M_\oplus$ and an impact angle $\theta=45^\circ$. We find
$$
\begin{align}
u_{T,\oplus} &= 30.813\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= -1.0285\;\text{km}\,\text{s}^{-1},\\
E' &= -411.87\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 2.1248\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.61109\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0788,\\
r_\text{p} = a'(1-e') &= 1.48411\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.73807\times 10^8\;\text{km},
\end{align}
$$
with $r_\text{p}$ and $r_\text{a}$ perihelion and aphelion. Evidently, the influence on the Earth's orbit is substantial.
In the case of a direct-from-behind collision, we get $\theta=0^\circ$, $v_\text{i}=11.2\,\text{km/s}$, so that
$$
\begin{align}
u_{T,\oplus} &= 30.803\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\
E' &= -412.72\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 2.1234\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.60778\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0695,\\
r_\text{p} = a'(1-e') &= 1.49598\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.71958\times 10^8\;\text{km}.
\end{align}
$$
As expected, the radius at impact has become the perihelion, and the change in eccentricity is lowest.
And just for fun, let's try the worst-case scenario: $\theta=180^\circ$, $v_\text{i}=72\,\text{km/s}$:
$$
\begin{align}
u_{T,\oplus} &= 23.239\;\text{km}\,\text{s}^{-1},\\
u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\
E' &= -617.10\;\text{km}^2\,\text{s}^{-2},\\
h'^2 &= 1.2086\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\
a' = -\frac{\mu}{2E'} &= 1.07530\times 10^8\;\text{km},\\
e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.391,\\
r_\text{p} = a'(1-e') &= 0.65462\times 10^8\;\text{km},\\
r_\text{a} = a'(1+e') &= 1.49598\times 10^8\;\text{km},
\end{align}
$$
so that the radius at impact has become the aphelion, and the change in eccentricity is highest. Although I wonder how much would be left of the Earth after such an apocalyptic event...
If the collision isn't central, then part of the energy will be transferred to the axial rotation of the Earth, which should reduce the effect on the orbit. But that will be more difficult to quantify.
Best Answer
A lot of debris has probably fallen back to earth. To stay in orbit you need enough angular momentum to overcome attraction. But if the collision happened at an angle a portion of the debris could have enough angular momentum to sustain orbit. Here is a nice video of how the collision could have happened.
Here are some snapshots from the video in case the link breaks
Over time any debris that had enough angular momentum to stay in orbit will eventually collect into the moon (or fall back to earth).
Edit based on the comments.
If you have some experience with orbital mechanics you might expect the debris to follow an ellipse. Since an ellipse forms a closed orbit which started at the surface you might expect that after one period all the debris would have fallen exactly where it came from. This isn't the case though: ellipsoidal curves only occur in two body systems. The blob of mass that is ejected is large enough that it has a gravitational field of its own and this complicates things a lot. Combine this with the fact that the blobs can collide with other blobs and stick to each other. This give some pretty complicated interactions and to say anything meaningful you would have to run a simulation at some point. In this case the interactions make some of the matter go into orbit instead of falling back down.