Hi,
In situations like this, there are two factors which need to be cosnsidered: are the control signals "Active High or "Active Low" and do the switches give "On/Off" or "High/Low" (Logic level) signals. An "Active High" (Positive Logic) AND gate is equivalent to an "Active Low" (Negative Logic) OR gate, and vice versa.
The outputs from a PICaxe will give true logic signals (i.e. approximately 0 and 5 volts) but a single transistor or relay contact will normally only "pull up" or "pull down" (depending whether it is connected to the supply rail or to earth). Some loads such as a filiament lamp or a "brushed" motor will pull the drive to the opposite logic level, but a LED is a diode which may have a significant forward voltage drop (several volts, above a typical "0" logic level, for blue and wite LEDs).
"Active Low" drive to higher powered devices (such as relays and LEDs connected directly to a positive supply rail) is often used (and may be preferable) but you have said that your LEDs are connected to ground. So they are driven "Active High", but by a transistor, which (only) pulls High. Therefore, the LED may be left to "float" when it is off, but adding a pull-down resistor for each "logic" input is probably all that's needed.
As mentioned above, the "two diodes" are probably your best solution, but it may be necessary to consider the factors above. The two diodes actually form an "Active Low" OR gate: i.e. if either input 1 OR input 2 is pulled low then the output must be low. Thus it can be considered as an "Active High" AND gate. But the diodes can only pull Low, so you may need a pull-up resistor to guarantee the "High" output voltage level.
For most Logic gates a weak pullup will be sufficient (>> 10 kohms), but the base drive to a bipolar npn transistor may need a lower resistance. The pullup current (or resistance) then determines how much pull-down needs to be provided through the diodes. Typically the pull-down resistor(s) should be several times smaller than the pull-up resistor, but normally still could be a mA or less.
Cheers, Alan.