Why is the first return map measure preserving?

This fact is general for any invertible measure preserving system:

Let $(X,\mathcal{B},\mu,T)$ be such that $T$ is invertible and measurable with $T^{-1}$ also measurable. Then $\mu(T^{-1}(A))=\mu(A)$ for all $A\in\mathcal{B}$ if and only if $\mu(TA) = \mu(A)$ for all $A\in\mathcal{B}$.

In fact we only need one direction of this theorem, that $\mu(TA)=\mu(A)$ for all $A\in\mathcal{B}$ implies that $\mu(T^{-1}(A))=\mu(A)$ for all $A\in\mathcal{B}$. This direction alone holds even if $T$ is not invertible, but is surjective. We will prove that in a minute, but this is what we need for your question.

Back to your question: Let $B$ be a measurable subset of $A$. Then since $T$ and $r_A$ are measurable, we have that $C:=T_A^{-1}(B)$ is also a measurable subset of $A$.

The proof given in the book shows that $\mu_A(T_A C) = \mu_A(C)$. Since $T_A$ is surjective (see proof below) we have $$T_A(C) = T_A T_A^{-1}(B)=B$$ which implies that $\mu_A(B)=\mu_A(T_A^{-1}(B))$ as required.

The proof that $T_A$ is surjective goes as follows: Consider the measure-preserving system $(X,\mathcal{B},\mu,T^{-1})$ and the same measurable set $A$. By poincare recurrence theorem for almost all $a\in A$ there exists $n$ such that $T^{-n}(a)\in A$. Let $b:=T^{-n}(a)$ for the minimal $n$, then $T_A(b)=a$. This shows that $T_A$ is essentially surjective which is enough for our purposes.