The only reason you have not vaporized your LEDs is because your AA batteries can't supply so much current. The "maximum 5V" you cite is the maximum reverse voltage these LEDs can before they break. Remember an LED is a diode which allows current in only one direction. 5V is the maximum voltage when you have them installed backwards
The forward voltage in the datasheet is 1.6V - 1.8V. This is, as you say, what you measured.
You should read How can I efficiently drive an LED? which covers most of this.
Firstly, you need to go back to school to learn the basics. What you are trying has no hope of working.
An LED is a current driven device, and it has a fixed forward voltage drop. Quite frankly, you are lucky the LEDs are even working at all any more - you must have LEDs that have a forward voltage at (or around) the 3V the batteries are supplying.
You have to run the LEDs with some form of current limiting device. At the moment that current limiting device is the batteries themselves being in the non-linear portion of the LEDs IV curve, which is very fortunate for you, or they would have died.
So your LEDs have a forward voltage of (around) 3V. Putting two in series makes that forward voltage 6V, so you will need at least 6V to drive them in series. You also need to limit that current to the right value to run the LEDs (no idea what that should be for your specific LEDs, but 20mA is common for normal LEDs).
So running from 9V would be sensible in this situation. Assuming 20mA and 3V forward voltage, you will need a resistor, of:
$$
V_{LED}=9V - 6V = 3V
$$
$$
R_{LED}=\frac{V_{LED}}{I_{LED}} = \frac{3}{0.02} = 150\Omega
$$
simulate this circuit – Schematic created using CircuitLab
You could also run from 4.5V (3xAA) and have as many LEDs in parallel as you wish. Each LED must have its own resistor in this case:
simulate this circuit
The resistor of 75Ω is worked out in the same way:
$$
V_{LED}=4.5V - 3V = 1.5V
$$
$$
R_{LED}=\frac{V_{LED}}{I_{LED}} = \frac{1.5}{0.02} = 75\Omega
$$
As a side note, it is more common to put the switch in the + side of the circuit, not the - side. It doesn't really make any difference, it's just that's where others would expect it to be who don't know the circuit you have designed.
Best Answer
Here is a straightforward Zener-based circuit which accounts for the forward voltage drop over the LEDs themselves:
simulate this circuit – Schematic created using CircuitLab
D1 is a 4.7V device, and D2 is an 11V device. These should give enough margin to work with different colours of LED, not just red (which tends to have the lowest forward voltage).
Note that since the full voltage is still applied to the first LED's limiting resistor, that will be the brightest one even when a higher LED is in fact lit. For that reason I do recommend that you use different coloured LEDs, taking advantage of the fact that red appears less bright than other colours.
An alternative circuit using transistors as LED drivers avoids this problem, and yields LEDs with identical brightness:
simulate this circuit
In this case, D1 is a 5.1V type and D2 is 10V. The precise values are less critical than with the first circuit; the forward voltage that has to be added to the Zener voltage to obtain the threshold is that of the transistor (typically 0.7V), not the LED.
Notice also that all three LEDs now draw from the same limiting resistor. This makes the brightness of each LED more nearly the same regardless of which battery is connected, since although the current through the resistor is increased with higher voltage, that current is shared by more LEDs. However, because the LEDs are now effectively connected in parallel, they should all be the same colour (and hence forward voltage).