Proving $(a+b)^* = b^*(ab^*)^*$ equationally
A report by Aceto contains an axiomatization of the equational theory of regular expressions, called "classical" by Conway (Table 1 on page 5). One of the axioms is very similar to what you're after: $(a+b)^* = (a^*b)^*a^*$. Using these axioms, we can derive your identity as follows (using associativity (C3, C10) for free):
$$
\begin{align*}
(a+b)^* &amp;= (b+a)^* &amp;&amp; C2 \\
&amp;= (b^*a)^*b^* &amp;&amp; C11 \\
&amp;= (\epsilon+b^*(ab^*)^*a)b^* &amp;&amp; C12 \\
&amp;= b^* + b^*(ab^*)^*ab^* &amp;&amp; C9 \\
&amp;= b^*(\epsilon + (ab^*)^*ab^*) &amp;&amp; C8 \\
&amp;= b^*(\epsilon + \epsilon(ab^*\epsilon)^*ab^*) &amp;&amp; C6,C7 \\
&amp;= b^*(\epsilon ab^*)^* &amp;&amp; C12 \\
&amp;= b^* (ab^*)^* &amp;&amp; C7
\end{align*}
$$