OK, let's try to walk through how this works. First of all, let's get rid of the capacitor at the output as that complicates things. Next, always remember that in op amp circuit (and most others for that matter) the best way to analyze what the op amp will do is to remember that it will swing its output to whatever it takes to force the negative input to be the same voltage as the positive input. If the output hits its maximum or minimum limit (its rails) while attempting to do this then this rule no longer applies. Ok, so much for basic op amp theory.
Now, imagine that the input is exactly at ground potential, that is the input voltage is exactly 0. This will be our starting point for analysis. Since the input is zero volts, and the op amp will swing its output to force its negative terminal to the same voltage as its positive terminal, we can assume for now that the negative input to the op amp is also at zero volts. By ohms law this means that there is no current flowing through R1. Since the negative terminal of the op amp has an infinite input resistance (at least for the sake of discussion), we can deduce that the current flow through R2 is also zero. So no current is flowing out of the op amp and through the diode, and the output of the circuit is 0 volts. We aren't sure of the output voltage of the op amp at this point other than we know it is not higher than approximately 0.5 or 0.6 volts and we are certain that no current is flowing through the diode, and we know for certain that the output of the circuit is 0 volts.
Next, let's imagine that the input starts to swing negative. At this point the op amp must do something to cause current to flow through R1 in order to insure that the negative input of the op amp remains at zero volts. The only way it can cause current flow from right to left is to force current flow through R2 from right to left. The only way that can happen is if the output of the circuit is starting to swing positive and away from zero. So, the output is going to track the input depending on the ratio of R2 to R1. Let's assume for now that R2=R1 to simplify things. In this case, the output of the circuit would exactly track the input except it would be inverted, that is swinging positive when the input is swinging negative. At this point we can also deduce that in order for the output to swing away from zero, the output of the op amp itself must be swinging above 0.6 volts. Aha! This is the essence of how this circuit gets rid of the diode voltage drop, so make sure you understand this point.
To continue, let the input voltage swing all the way negative and then come back to zero. All the while, the output of the circuit has tracked this volt for volt except the output is positive while the input is negative.
Now, what happens when the input swings positive? Ah, now things work very differently. The moment this swing begins, current flow through R1 starts to go from left to right. So, current flow through R2 must go left to right. The trouble is that this is impossible because the current can't pass through that diode from anode to cathode (i.e right to left). So the op amp output tries to go below 0.6 volts, but cannot get current to flow in the correct direction through R1 and R2 to get the negative terminal to zero volts. But it tries as hard as it can by swinging to its negative rail. It can't go any further so it just sits at that rail. Meanwhile, no current flows through the diode from right to left. You see what has happened now? The op amp is actually disconnected from the output of the circuit because of the diode. Now, if you were to place a resistor from the output of the circuit to ground, you would see that the input voltage would be divided by the values of R1, R2 and the output load resistance. In other words, the input voltage would cause current to flow through R1 from left to right, through R2 from left to right and through the load resistor. Now, if the load resistor is a very high resistance compared to R1 and R2, almost all of the input voltage would be impressed across the load resistor.
Do you see what has happened? A positive swing of voltage at the input shows up directly across the load resistor. And yet, we deduced that a negative voltage swing at the input also shows up as a positive voltage swing at the output. Isnt' that the definition of a full wave rectification? So, the answer is that this circuit does indeed provide full wave rectification if the output load resistance is much higher than R1+R2. And this circuit does indeed eliminate the diode drop.
That's it for the explanation except to mention that once you fully understand it, try putting the capacitor back accross the output and then tell us what that does.