Proof.
Consider the map of operads \([1] \otimes \mathbb E_1 \rightarrow\mathcal O\) representing the \(\mathbb E_1\)-algebra map \(f \colon A \rightarrow B\). The fiber of \(\mathrm{PreBraid}_{\mathcal O}(f) \rightarrow\mathrm{PreBraid}_{\mathcal P}(F(f))\) at a prebraiding represented by an operad map \(\mathbb T_2 \otimes \mathbb E_1 \rightarrow\mathcal P\) is precisely the space of (dashed) lifts of the following commuting square of operads: 
Since \([1] \otimes \mathbb E_1 \rightarrow\mathbb T_2 \otimes \mathbb E_1\) is \(0\)-surjective, it follows from lemma 8.3.3 that this space of lifts is equivalent to the space of lifts 
Hence, replacing \(\mathcal O\) by \(\mathcal O|_{[1] \otimes \mathbb E_1}\) and \(\mathcal P\) by \(\mathcal P|_{[1] \otimes \mathbb E_1}\) with multimapping spaces as in observation 8.3.2, we may without loss of generality assume that the maps \[\mathrm{Mul}_{\mathcal O}(a, \ldots, a; b) \rightarrow\mathrm{Mul}_{\mathcal P}(Fa, \ldots, Fa; Fb)\] are equivalences. It then follows from proposition 7.8.2.([00GH]) that the maps \[\mathrm{Mul}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal O)}(a, \ldots, a; b) \rightarrow\mathrm{Mul}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal P)}(Fa, \ldots, Fa; Fb)\] are equivalences. Hence, \[\begin{aligned}
\mathrm{PreBraid}_{\mathcal O}(g) = &\mathrm{Mul}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal O)}(a,a; b) \times_{\mathrm{Mul}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal O)}(a, b)^{2}} \{g\} \\
&\longrightarrow \mathrm{Mul}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal P)}(Fa,Fa; Fb) \times_{\mathrm{Mul}_{\mathrm{Alg}_{\mathbb E_1}(\mathcal P)}(Fa, Fb)^{2}} \{Fg\} = \mathrm{PreBraid}_{\mathcal O}(Fg)
\end{aligned}\] is an equivalence. ◻