I have done some more research about this question and here is my (partial) conclusion.
The theorem as I stated it in the body of the question seems to be wrong because it's too general. What seems to be known is essentially just what is proved in the paper of Lehmann and Scheffé linked by Iosif Pinelis (where they get a possibly non-measurable $\phi$), and the obvious corollary that can be obtained from it for the measurable case, which I state below.
If a set function $S:\mathcal{X}\to E$ satisfies condition $(\star)$, if $L^1(\mathcal{X})$ is separable, and if we endow $E$ with the final $\sigma$-algebra $\mathscr{E}$ ($A\in \mathscr{E}$ if and only if $S^{-1}(A)\in\mathscr{X}$), then $S:(\mathcal{X},\mathscr{X})\to (E,\mathscr{E})$ is a measurable sufficient statistic which is "measurably minimal" in the class of all measurable sufficient statistics $T$ with the final $\sigma$-algebra on the codomain, that is $S=\phi(T)$ for some measurable $\phi$.
I haven't found anything about the general case where the codomains are endowed with arbitrary $\sigma$-algebras and hence I suppose that in this case the result fails (but haven't found any counterexample). In facts, after consulting some of the main references in the field (1, 2, 3) I got somehow convinced that the right way of thinking of minimal sufficient statistics in this theorem is in terms of set functions, not of measurable functions. I have also found that the concept of minimal sufficient $\sigma$-algebra, which is to be the standard approach to minimality in the most recent papers, is equivalent to the one of "measurably" minimal statistics on Polish sample spaces (see Theorem 4 in Rogge - and Landers and Rogge for a counterexample when the space is not Polish).