Yes! Here is a proof which is slightly different from both your proof and the proof in Mines–Richman–Ruitenberg.
First define the similarity relation on $\mathrm{List}(R \times X)$ as in Mines–Richman–Ruitenberg as the smallest equivalence relation generated by their clauses (i), (ii) and (iii). I'd argue that this is the obvious thing to do if you set out to construct the free module.
We'll then show: Lists $[\langle a_1, x_1\rangle, \ldots, \langle a_n, x_n \rangle]$ and $[\langle b_1, y_1\rangle, \ldots, \langle b_m, y_m \rangle]$ are similar if and only if for all $x \in X$, $I \subseteq \{1,\ldots,n\}$, $J \subseteq \{1,\ldots,m\}$ (I just mean the detachable subsets here) such that $x_i = x$ for all $i \in I$ and $y_j = x$ for all $j \in J$, there are supersets $I' \supseteq I$ and $J' \supseteq J$ (again, $I'$ and $J'$ should be detachable) such that $x_i = x$ for all $i \in I'$, $y_j = x$ for all $j \in J'$ and $\sum_{i \in I'} a_i = \sum_{j \in J'} b_j$.
This is done in three steps:
- The proposed characterization of similarity defines an equivalence relation (transitivity is slightly nontrivial, as in your proof).
- That equivalence relation encompasses similarity.
- It is the smallest equivalence relation satisfying (i), (ii) and (iii). For this step to work, it's important that the additive group of the ring is (as all groups are) cancellative. That is, this step wouldn't generalize to rigs $(R,0,1,+,\cdot)$ where $(R,0,+)$ is just required to be a commutative monoid.
With the characterization of similarity in hand, the injectivity of the unit is then easy to establish: Assume that the lists $[\langle 1, x_1\rangle]$ and $[\langle 1, y_1\rangle]$ are similar. By the characterization (applied to $x_1$, $I = \{1\}$, $J = \emptyset$), there is a detachable set $J' \subseteq \{1\}$ such that $y_j = x_1$ for all $j \in J'$ and $1 = \sum_{j \in J'} 1$. We have $1 \in J'$ or $1 \not\in J'$. If $1 \not\in J'$, then $1 = 0$ in $R$; a contradiction. Thus $1 \in J'$. So $y_1 = x_1$.