By Cauchy identity, $${}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1,$$ we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=(1-z)^{-a-b},$ $${}_1\phi_0(a;—;q,z){}_1\phi_0(b;—;q,az)={}_1\phi_0(ab;—;q,z),$$ which is $$\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n\sum_{m\geq0}\frac{(b;q)_m}{(q;q)_m}(az)^m=\sum_{n\geq0}\frac{(ab;q)_n}{(q;q)_n}z^n.$$

Comparing the coefficients of $z$ in the both side of equation above, the identity in the title can be established as $$\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}.$$

However, what I really concern is that there is any possible for us to prove this identity directly without utilizing Cauchy identity. Maybe we can use **mathematical induction** on *$n$* or some other **combinatorial** or **algebraic** proofs?