Capacitors as a Storage for Electricity
Yes, it will light the LED up but for a very short time.

The amount of energy stored by a capacitor is:

\$E=\frac{1}{2} C V^2\$

E is the energy stored, C is the capacitance of the capacitor, and V is the voltage across the capacitor.

For a 1uF capacitor charged to 9V, that is 40.5 uJ.

In addition to this, typical LED's have a forward current drop of ~2-3V, and capacitors typically have lower internal resistances compared to batteries. This means that you're going to get a very high current for a very short time, which could damage your LED.

Suppose you did limit the output current to 10mA (typical for small LED's), and the LED has a forward voltage drop of ~2V.

That means the LED would stay "lit" for (assuming we could extract every last bit of energy at the same rate from the capacitor):

\$
\frac{40.5uJ}{2V \cdot 10 mA} = 2.025 ms
\$

Which is a very short time indeed.