Attenuation and Insertion Loss for a 2 port network
Note: you have your notation backwards from what I believe to be convention. \$b\$ refers to the outgoing wave from the network and \$a\$ refers to the wave incident on the network. That being said, the following will use your notation instead.



When \$\Gamma_g=0\$, what is implied is that the source impedance of your generator at port 1 is matched to the port's impedance. This means that any wave coming out of port 1 will be completely absorbed by the source's impedance and no wave will be reflected back into the network. However, the source is generating it's own wave which is incident on the network. Thus, the total wave incident on port 1 of the network would be 
$$ b_g=\Gamma_g a_g + V_g = 0 + V_g = V_g$$
where \$V_g\$ is the voltage wave generated by the source. \$b_g\$ is not zero, like you mentioned. It is this voltage wave that is transmitted, and against which you calculated the attenuation.

You have to remember that the S parameters of a network and defined independent from what's connected to the network. The only thing you need to know is each port's reference impedances (which are usually all taken to be \$50\Omega\$ in standard microwave systems). Attaching a source to port 1 requires that you include the source's voltage in \$b_g\$. The equation involving \$a_L\$ would now read (assuming matched ports)
$$ a_L = S_{21} b_g + S_{22} b_L = S_{21} V_g + S_{22}*0 = S_{21} V_g $$
Then the attenuation according to your definition is 
$$ \frac{|V_g|^2}{|a_L|^2} = \frac{1}{|S_{21}|^2} $$

Note again that this answer assumes both load and source are matched to the same reference impedance of the network.