= Solution
Retain the specified leading poles and write $A=a_1/\epsilon$, $B=b_1/\epsilon^2$. Solving $g=g_0+Ag_0^2+Bg_0^3$ recursively gives
$$
g_0=g-Ag^2+(2A^2-B)g^3+O(g^4).
$$
The scale derivative must again be reexpressed in the <renormalized coupling>:
$$
\begin{aligned}
\beta(g)&=-\epsilon g_0\left(1+2Ag_0+3Bg_0^2\right)\\
&=-\epsilon g-\epsilon A g^2+2\epsilon(A^2-B)g^3+O(g^4)\\
&=\boxed{-\epsilon g-a_1g^2+\frac{2(a_1^2-b_1)}{\epsilon}g^3+O(g^4).}
\end{aligned}
$$
The <leading-pole consistency under coupling reparametrization> shows that a finite <renormalization-group beta function> requires the two-loop double-pole consistency condition
$$
\boxed{b_1=a_1^2.}
$$
The leading two-loop pole is fixed by the one-loop simple pole; it is not an independent parameter. This is an instance of the <simple-pole formulas for minimal-subtraction renormalization group functions> and their higher-pole consistency relations.
The source's leading-pole truncation does \b[not] fix the finite two-loop coefficient of the <beta function>. For example, adding the omitted term $c_1g_0^3/\epsilon$ leaves the double-pole condition unchanged but contributes $-2c_1g^3$ to the finite <beta function>. Thus cancelling the displayed $1/\epsilon$ term is not a claim that the physical two-loop running vanishes.
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