closed loop pole of negative feedback
The forward path TF, including a pole, is
$$\small G_1 (s)=\frac{a}{(1+\tau s)}=\frac{a/\tau}{s+1/\tau}=\frac{a/\tau}{s+\omega_{B1}}$$
where \$\small \omega_{B1}\$ is the bandwidth.
Thus, the pole is at \$\small s=-1/\tau =-\omega_{B1}\$, the DC gain is \$a\$, and the gain-bandwidth product, is $$\small GBW= \frac{a}{\tau}$$
The closed loop TF is: $$\small G_2 (s)=\frac{a/\tau}{s+(1 +af)/\tau}=\frac{a/\tau}{s+\omega_{B2}}$$
Hence, the pole has moved to \$\small s=-(1 +af)/\tau=-\omega_{B2}\$, and the bandwidth has increased by a factor, \$\small (1+af)\$.
the DC gain is now: \$\small a/(1+af)\$, so the gain-bandwidth product remains at: \$\small GBW=\large \frac{a}{\tau}\$, as, of course, it must.