Maybe a non-rigorous answer formulated in the language of perturbative QFT helps.

Examples can be produced in this context by performing non-trivial redefinitions of the fields. The off-shell Green functions in the original QFT and in the one with redefined fields will not be equal in general, but the S-matrices will, by the equivalence theorem.

The simplest explicit example I can think of is the two Lagrangians: $$ \mathcal{L}_1 = (\partial_\mu \phi)^2, \qquad \mathcal{L}_2 = (\partial_\mu \phi)^2 - 2 \lambda \phi^2 \partial^2 \phi - \frac{4\lambda^2}{3} \phi^3 \partial^2 \phi. $$ related by $\phi \to \phi + \lambda \phi^2$. Consider for example the 4-point function for $\mathcal{L}_2$. It receives contributions from both the second and third term, and it is not equivalent to the trivial one generated by $\mathcal{L}_1$. However, it can be checked that it takes the form $$ G(p_1, p_2, p_3, p_4) = \frac{p_1^2 + p_2^2 + p_3^2 + p_4^2}{p_1^2 \,p_2^2 \, p_3^2 \, p_4^2}f(p_1, p_2, p_3, p_4), $$ with $f$ an analytic function at $p^2_i \to 0$, so that the corresponding on-shell amplitude vanishes. In this case, it does not matter that the coefficients of the two interaction terms are correlated, but it will for other Green functions. Of course, the equivalence theorem will still hold, but it will become more and more difficult to check it explicitly.

Maybe a non-rigorous answer formulated in the language of perturbative QFT helps.

Examples can be produced in this context by performing non-trivial redefinitions of the fields. The off-shell Green functions in the original QFT and in the one with redefined fields will not be equal in general, but the S-matrices will, by the equivalence theorem.

The simplest explicit example I can think of is the two Lagrangians: $$ \mathcal{L}_1 = (\partial_\mu \phi)^2, \qquad \mathcal{L}_2 = (\partial_\mu \phi)^2 - 2 \lambda \phi^2 \partial^2 \phi - \frac{4\lambda^2}{3} \phi^3 \partial^2 \phi. $$ related by $\phi \to \phi + \lambda \phi^2$. Consider for example the 4-point function for $\mathcal{L}_2$. It receives contributions from both the second and third term, and it is not equivalent to the trivial one generated by $\mathcal{L}_1$. However, it can be checked that (at tree level) it takes the form $$ G(p_1, p_2, p_3, p_4) = \frac{p_1^2 + p_2^2 + p_3^2 + p_4^2}{p_1^2 \,p_2^2 \, p_3^2 \, p_4^2}f(p_1, p_2, p_3, p_4), $$ with $f$ an analytic function at $p^2_i \to 0$, so that the corresponding on-shell amplitude vanishes. In this case, it does not matter that the coefficients of the two interaction terms are correlated, but it will for other Green functions. Of course, the equivalence theorem will still hold, but it will become more and more difficult to check it explicitly.