1) When you see + and - in the datasheet for a relay, it is always the polarity you should use from your power supply.

2) If we ignore any internal zener and just assume you have a "normal" relay with an internal diode, then it is important that you don't try to feed the relay with reversed polarity. The diode-will short-circuit if you reverse the polairty, and unless the power supply has a current-limitation, you will destroy the diode.

3) You are switching the relay on the negative side. The positive side of the relay is always at +5V. When you disconnect the ground side of the relay, the relay will continue to drive a current in the same direction as before. Without a diode, this current will go through the air, or through other electronic components. With the free wheeling diode, the current will just loopback between the two poles of the relay coil, and the voltage over the relay coil will then be one diode voltage. But since your + pin of the relay is already fixed at +5V, your - pin of the relay will be one diode voltage higher than +5V.

http://en.wikipedia.org/wiki/Flyback_diode

All DC-driven inductive loads needs a free wheeling diode to protect the circuit when you disconnect the load. In some situations, a diode exists inside the relay. This will require that you drive it with DC of correct polarity. In some situations, the switching transistor may have a design allowing it to swallow the energy. But sutch a switching transistor will only protect itself (from a limited number of load disconnects/second) - it will not protect other equipment (such as your LED) since any protection inside the transistor may allow a quite high voltage compared to the normal coil voltage.

It was bad of the relay manufacturer to not show a schematic explicitly presenting their diodes. With a 56V zener in series with a normal diode, you will have to treat the relay as not having any internal protection and instead use an external diode. Or find another relay manufacturer. The important thing is that you need a diode design that will not let the relay coil build up a high voltage at disconnect.

Don't be in such a hurry to experiment. Do read the answers you get.

Note that you may design without the diodes, and if all you do is design by experiment, it may look like it works. But what your experiments may not show, is how the 7407 buffer chip and the LED will age by the abuse. Your experiments will not notice this unless you get a total failure. This may happen on the first disconnect. But it may just happen after 100 or 1000 disconnects. That is why you always have to design based on a theoretic foundation. It isn't enough if something looks correct.