Does a BJT amplify the biased DC voltage or just the AC/DC input voltage?
Amplification, by definition, is associated with change in the voltage (or current or whatever parameter involved). So when you say that the voltage gain of the amplifier is say 10, you mean that a change in the input voltage by a certain amount would result in 10 times change in the output voltage. So, in fact, the formula you have quoted in the question is the ratio of the change in the output with change in the input voltage:
$$\frac{dV_o}{dV_i} = -\frac{R_L}{R_E + R_e}$$
So actually amplification of the bias voltage itself does not make sense because there are no changes involved in that voltage.
So, suppose you are getting a gain of 10 with input bias of 0.7V and output at 4V. What your equation means is that if you increment the input bias (at 0.7V) by 0.1V (just as example) the output bias will go down by 10*0.1 = 1V from 4V.
But, be aware, that it is just the small signal gain of your circuit. So if you swing the input bias too much then you will not get the 10x change at the output as before (but a little less).