Free splittings of one-relator groups

I think Grushko plus the Freiheitssatz does the trick. Suppose that $G=A\ast B$ is a 1-relator group which splits as a free product non-trivially. By Grushko, $rank(G)=rank(A)+rank(B)=m+n$, where $rank(A)=m, rank(B)=n$. If $G$ is not free, then by Grushko there is a 1-relator presentation $\langle x_1,\ldots, x_m ,y_1,\ldots, y_n | R\rangle$, such that $\langle x_1,\ldots, x_m \rangle =A \leq G, \langle y_1, \ldots, y_n \rangle=B \leq G$ (for this, one has to use the strong version of Grushko that any 1-relator presentation is Nielsen equivalent to one of this type). Suppose that $R$ is cyclically reduced, and involves a generator $x_1 \in F_m$. Then $\langle x_2,\ldots , x_m, y_1, \ldots, y_n\rangle$ generates a free subgroup of $G$ by the Freiheitssatz. But this implies that $B = \langle y_, \ldots, y_n\rangle$ is free. Moreover, if $R$ involves one of the generators $y_i$, then one sees that $A=\langle x_1,\ldots,x_m\rangle$ is also free, and therefore $G=A\ast B$ is free, a contradiction. So $R \in F_m$, as required.

I just came across the following, which is Prop. II.5.13 of Lyndon-Schupp.

Proposition. Let $G = \langle x_1, \ldots , x_n: r\rangle$ where $r$ is of minimal length under $Aut(\langle x_1, \ldots , x_n\rangle)$ and contains exactly the generators $x_1,\ldots , x_k$ for some $0 \leq k \leq n$. Then $G \cong G_1*G_2$ where $G_1 = \langle x_1, \ldots , x_k:r\rangle$ is freely indecomposable and $G_2$ is free with basis $x_{k+1}, \ldots ,x_n$.

Unless I'm missing something, up to isomorphism you can assume that your relator has minimal length. If it is not contained in a free factor of $G$, then $k=n$ in the Proposition, hence $G = G_1$ is freely indecomposable.

Your question is reminiscent of Jaco's lemma. A special case of Jaco's lemma applies to a 2-handle attached to the boundary of a (3-dimensional) handlebody $H$ (which has free fundamental group). If the boundary curve $J\subset \partial H$ along which the handle is attached is "disk-busting", that is, $\partial H-J$ is incompressible (and therefore $\pi_1$-injective) in $H$, then the manifold obtained has $\pi_1$-injective boundary (and therefore does not split as a free product). This condition is easily seen to be equivalent to the conjugacy class of $J$ not belonging to any free factor of $\pi_1(H)$. So this answers your question in this very special case (also note that this works for "orbifold" handles attached along $J$). It's not clear whether Jaco's method might apply in your case, but it might be worth having a look (there are other proofs and generalizations of it too which you can find through Mathscinet). In particular, his argument might also apply if the word is only virtually geometric.