To the best of my understanding, the universal property of the polynomial ring can be phrased (see page 5 here):
Given a commutative and unital $R$, then the polynomial ring $R[X]$ satisfies the universal property that for any pair $(\phi, s)$, where $\phi: R \to S$ is a ring homomorphism, $s \in S$, and $S$ is also a commutative unital ring, there is a unique ring homomorphism $\Phi: R[X] \to S$ such that $\Phi(x)=s$ and $\Phi \circ \iota = \phi$, where $\iota: R \to R[X]$ is the standard inclusion.
Note that I assume ring homomorphisms preserve the multiplicative identity.
Question: Is the following a correct equivalent formulation?
Given a commutative unital ring $R$, the polynomial ring $R[X]$ satisfies the universal property that for any pair $(\phi, s)$, where $\phi: R \to S$ is a ring homomorphism and $s \in S$ is such that $\sigma \cdot s = s \cdot \sigma$ for all $\sigma \in \operatorname{Im}(\phi)$, and $S$ is a unital ring (not necessarily commutative), then there is a unique ring homomorphism $\Phi: R[X] \to S$ such that $\Phi(x) =s$ and $\Phi \circ \iota = \phi$, where $\iota: R \to R[X]$ is the standard inclusion.
Motivation: The one-to-one correspondence between (i) $R[X]$ module structures on a given abelian group $M$ compatible with a given $R$-module structure on $M$ and (ii) (abelian) group endomorphisms on $M$ which are $R$-linear with respect to the given $R$-module structure, would be a direct consequence of the second formulation. (Compare page 73 here.)
Take $S = \operatorname{End}(M)$, $\phi$ the $R$-scalar multiplication, $\Phi$ the $R[X]$-scalar multiplication, and $s$ to be the $R$-linear (w.r.t. $\phi$) endomorphism of $M$. ($s$ commuting with all $\sigma \in \operatorname{Im}(\phi)$ is what makes it $R$-linear.)
Proof Attempt: I'll skip details for brevity, but basically I think I was able to show that an $s \in S$ commutes with all $\sigma \in \operatorname{Im}(\phi)$ if and only if $s$ is contained in some commutative subring of $S$ which also contains $\operatorname{Im}(\phi)$.
It should hopefully also be true that every such subring contains the subring generated by $s$ and $\operatorname{Im}(\phi)$ (called $k[f]$ on p.73 of the linked notes), and so the $\Phi$ we get by applying the standard formulation of the universal property of the polynomial ring should be the same regardless of which commutative subring of $S$ containing both $s$ and $\operatorname{Im}(\phi)$ we choose.
The other direction is trivial since obviously any $s \in S$ will satisfy the required condition when $S$ is itself commutative.
Best Answer
Yes, this works. The proof is basically identical to the proof for commutative rings, since the assumption that $\sigma s=s\sigma$ for all $\sigma \in \operatorname{Im}(\phi)$ means that all the elements of $S$ you will ever write down in the proof commute. Or alternatively, as you mentioned, you can just say that the subring generated by $s$ and $\operatorname{Im}(\phi)$ is commutative and then apply the result for commutative rings.