Let $b$ be the inverse of $a$, then for the translation term $T$(Vector), rotation term $Q$(Quaternion), scale term $S$(Vector):
$$\begin{equation}
\begin{split}
T_(a * b) & = T_b + (Q_b \times (S_b \times T_a) \times Q_b^{-1}) = 0 \\\
T(b) & = -(Q_b \times S_b \times T_a \times Q_b^{-1}) \\\
& = Q_b \times (S_b \times -T_a) \times Q_b^{-1}
\end{split}
\end{equation}$$
Which equals to:
- Translate by $-T_a$
- Then scale by the $S_b$(ie. $S_a^{-1}$)
- Then rotate by $Q_b$(ie. $Q_a^{-1}$)