Just a few nit picky comments on your schematic; I would first recommend that you add a resistor between the diode D1 and the base of Q1 in order to protect Q1 since it's a current controlled device and a diode does not limit current that much. Second, I would add a flyback diode across the coil of the relay for protection. Correct me if I'm wrong, but what you want to do is connect the circuit so that when you shift out of neutral, the neutral light goes out, then comes back on once you shift back to neutral?
Unfortunately there isn't really any other way to go about connecting the circuit as you described it, reason is the light bulb has a high current draw. A relay can easily handle this current, but there aren't a lot of other parts that can. Most general purpose NPN or PNP transistors are rated for 500mA or less. A 12V 10W light bulb will draw at least 800mA. The few parts that can handle the current for such a high current is usually a power transistor or some sort of power MOSFET. But then the heat sink for these parts would need to be large enough to dissipate the heat appropriately, which would make them even larger than the original relay.
However, if you are planning to use LEDs instead of a light bulb, that changes everything about this circuit. You no longer need a relay, and you can even use general purpose transistors which are 1/8 of a cubic inch. you can even make one LED display for both (using a bi-color LED).
I have attached some circuits that should clarify what I am talking about. The first one is a simplified circuit of what you have now. The second one is of a circuit that uses power MOSFETs to drive the light bulbs, remember, the heat sink needs to be able to disipate 10W of power or more for a 10W bulb. The third is of a simple circuit if you are using LEDs instead of a light bulb. I prefer to use the MOSFET or the LED approach over a relay because they physically last longer. Hope this helps
(P.S. Your circuit is latching the relay, remove D2 and R2)