If you settle for a noncontracting (or monotonic) grammar instead:

$S\to DTA \mid ab$ $~~~~~$ the last $A$ is an end-marker for the last $ab^n$, $n=1$ separate.

$T\to DTa \mid Da$ $~~~~~$ number of $D$'s equals the number of $a$'s ($A$ counts as $a$)

$Da \to abD$ $~~~~~$ every $a$ gets an extra $b$

$Db \to bD$ $~~~~~$ $D$ moves over $b$'s

$DA \to Ab$ $~~~~~$ and $D$ disappears at the last block of $ab^n$

$A \to a$ $~~~~~$ finally end-marker $A$ is changed into $a$ (if we do this too soon, the $D$'s will not disappear, and the dirivation is not valid)