I am just posting as an answer my comments above, so that this question will not remain unanswered.

One case where the morphism $f$ is non-immersive, in fact non-injective, is when the special fiber $X_s$ contains an irreducible component $Y$ that is a genus $0$ curve such that every connected component $Z_1,\dots,Z_m$ of the residual curve $Z$ to $Y$ in $X_s$ intersects $Y$ in a single ordinary double point $q_i$. For this to be stable, $m$ must be at least $3$. Also, the sum of the arithmetic genera, $p_a(Z_1)+\dots +p_a(Z_m)$, equals the arithmetic genus $p_a(X_\eta)$ of the generic fiber $X_\eta$.

For simplicity, assume that every curve $Z_i$ is a curve of compact type, so that the "Jacobian" of $Z_i$ is a product of Abelian varieties that has dimension equal to $p_a(Z_i)$. Then the Neron model of the Jacobian $\text{Jac}(X_\eta)$ of the generic fiber is an Abelian scheme over $S$. Since all morphisms from a genus $0$ curve to an Abelian variety are constant, $f$ must be constant on $Y\setminus \{q_1,\dots,q_r\}$. Thus $f$ is not an immersion; it is not even injective.

This seems to contradict the OP's assertion that $f$ is an immersion whenever the scheme-theoretic image of $f$ has a reduced special fiber. It is straightforward to arrange a special fiber as above and such that $X$ is regular. By the Weil extension theorem, the rational transformation $f$ extends to a regular morphism on all of $X$, not just on $X\setminus X_s^{\text{Sing}}$. This morphism necessarily factors through the contraction $X'$ of $Y$ in $X$. To prove that the induced morphism is a closed immersion, it suffices to prove that the morphism is unramified at the image point $y'$ of $Y$ in $X'$. The Zariski tangent spaces of the curves $Z_i$ at $q_i$ together with the Zariski tangent space of a transversal give a direct sum decomposition of the Zariski tangent space to $X'$ at $y'$. It seems to me that these subspaces map to linearly independent subspaces of the Zariski tangent space of $J_X$, so that the morphism is unramified. Thus the special fiber of the scheme theoretic image of $f$ equals the special fiber of $X'$. Since $X'$ is normal, the special fiber is $S1$. Since the special fiber is generically reduced, it is everywhere reduced.