Before i start my ramblings, you might want to take a look at the Brammo Empulse. It's an electric motorcycle with a gearbox and a clutch, and it may give you some interesting information. Take a look here.
In your case, you can't really do without a gearbox like commented.
You could do without a gearbox but only if your powersupply and engine combination allows so, and if you don't care about getting less miles per kWh. A Tesla S doesn't have a gearbox for instance. You can take a motor and a fixed gear ratio that can give you the tractive force that you want to attain, and you just feed more and more voltage to the motor as your speed increases, as long as you're under the max current and rpm of the motor, you can feed the motor as much voltage as you want. It's not very efficient any more at those speeds though. That's why a Tesla has a relatively low top speed despite it's great motor power. I've driven one, and you can deplete a Tesla's battery in minutes when you demand full power. An electric motor's efficiency gets quickly worse when you demand torque, while a combustion engine's efficiency increases. Comparing combustion engines with electric motors is comparing apples with pears, as per the saying in my language.
Your problem is that your powersupply probably has a relatively low voltage, so you can't pull off this trick. The Brammo can't do either, its battery only supplies 103V, while a Tesla supplies 400V. To determine the best gear ratios for your motor, you have to know its efficiency curve, to make maximum use of your motor's powerband. Especially because you're heavily underpowered, with 300W.
To get meaningful and effective ratios, I would set up goals for the steepest hill you want to be able to get up, and the maximum speed you want to achieve. With the weight of you and your bike, you can calculate the tractive force you'll need to have for these goals. Then you can calculate the transmission ratios you need for that.
(hereunder follows an awfully bad explained calculation of which i'm sure and hope someone will give a better version of)
Let's say the combination of you and your bike weigh 100kg together, thats 981 Newton. You want to get up a hill of 10% or 6degrees, not abnormally steep. This gives a sloperesistance of (F=sin[a]x m x g) sin(6deg)x 981= 103 Newton. If you can fully use your 300W this gives you a speed of (P=F*v) 300/103= 2.9m/s or 10.5km/h, if you neglect rolling friction. That's not really fast but still reasonable. To get a tractive force of 103N you'll have to multiply this by your wheel radius to get wheel torque. Suppose your wheel radius is 35cm. In that case it's (M=Fxa) 103x0.35= 35.9Nm of wheel torque. Divide that by your motor torque, and you have your ratio.
Suppose your 300W motor has a maximum rpm of 8000, and a torque of 0.359Nm.
Then you need a total ratio of around 100:1 for first gear: 35.9/0.359= 100. On a flat road, this tractive force gives you an acceleration of 103/100= 1.03m/s2 until your motor reaches maximum rpm, or the current decreases due to increasing back-emf. That's a reasonable acceleration. Suppose your motor has a maximum rpm of 8000, which is 838rad/s. That means that your wheel is spinning at 838/100= 8.38rad/s because of your transmission ratio. The radius was 0.35m so your speed is (v=w*r) 8.38x0.35= 2.9m/s or 10.6km/h. That means there's need for more gears. Let's say you want to get to 18km/h in 2nd gear. And 30km/h in 3rd gear. That means you need ratios of 18/10.6x100= 59:1 and 30/10.6x100= 35:1.
These ratios will give you accelerations of 59/100*1.03 = 0.6m/s2 and 35/100*1.03= 0.36m/s2. If you take a final drive of 1:100 you get ratios of 1.72 and 2.85, and a prise direct/direct drive.
These calculations don't take friction in account, but it's just for an impression. If i have time i'll improve this post a bit.