The devil is in the details. When you say that

...modulo formal issues involving multiplying operator-valued distributions

right after you wrote the equations of motion you ignore a very important point. The right-hand side of the equation

$$ (\partial^2 + m^2) \hat{\phi} = \hat{\phi}^3 $$

is not well defined if you does not regularize and renormalize it. Turns out that in other to do it exactly you need in advance the complete (quantum) solution of the model. One way to circumvent this problem is by organize your computations by perturbation theory of the non-linearity. Then at each order in your computation, for a given regularization, you are going to need to introduce a counter-term to match some renormalization conditions (and avoid nonsensical divergences). The anomaly are going to appear by the fact that there is no regularization, and no finite conter-term, able to preserve the anomalous symmetry. This imply that any attempt to make sense of the equation of motion you wrote are going to break the classical symmetry.

This conclusion does not depends on the perturbation theory. You may try to make sense of this equation by putting it on the lattice use computers to do non-perturbative computations. By putting the theory on the lattice you are going to regularize the theory first. The anomaly will appear as an impossibility to put this theory on the computer without breaking (explicitly) the symmetry.

A more abstract way to see what goes wrong. In free theory operators like $\phi^3$ need to be normal ordered. Once you introduce an interaction, the normal ordering are going to be modified, and you can calculate this modifications perturbativally. The calculation is obtained by looking at $\phi^3$ as a vertex of Feynman diagrams, and the normal ordering of free theory as self-contraction of the legs that come out of the vertex. In the interacted case there will be also collision of this legs, with new vertices appearing on this collisions.

About the last equation, this is an identity between operators in the Heisenberg picture. Again, classically you expect that the left-hand side is identically zero as an operator equation. The fact that the current operator suffers from ordering problems in the case of chiral fermions, you are going to get a correction. What is interesting is that this correction is exact at one-loop computation. Usually equations like the first one you wrote will continue to get corrections indefinitely. You can think about this equations already corrected as coming from the 1PI quantum effective action.

So, in your example of the chiral anomally, the 1PI quantum effective action at one-loop will have a term that breaks the chiral anomally and contribute with the equations of motion.

The devil is in the details. When you say that

...modulo formal issues involving multiplying operator-valued distributions

right after you wrote the equations of motion you ignore a very important point. The right-hand side of the equation

$$ (\partial^2 + m^2) \hat{\phi} = \hat{\phi}^3 $$

is not well defined if you does not regularize and renormalize it. Turns out that in other to do it exactly you need in advance the complete (quantum) solution of the model. One way to circumvent this problem is by organize your computations by perturbation theory of the non-linearity. Then at each order in your computation, for a given regularization, you are going to need to introduce a counter-term to match some renormalization conditions (and avoid nonsensical divergences). The anomaly are going to appear by the fact that there is no regularization, and no finite conter-term, able to preserve the anomalous symmetry. This imply that any attempt to make sense of the equation of motion you wrote are going to break the classical symmetry.

This conclusion does not depends on the perturbation theory. You may try to make sense of this equation by putting it on the lattice and use computers to do non-perturbative computations. By putting the theory on the lattice you are regularizing the theory. The anomaly will appear as an impossibility to put this theory on the computer without breaking (explicitly) the symmetry.

A more abstract way to see what goes wrong is to work with renormalized operators. In free theory, operators like $\phi^3$ need to be normal ordered (renormalized). Once you introduce an interaction, the normal ordering are going to be modified, and you can calculate this modifications perturbativally. This calculation is obtained by looking at $\phi^3$ as a vertex of Feynman diagrams. The normal ordering of free theory is the subtraction of self-contraction of the legs that come out of the vertex. In the interacted case there will be also collision of this legs, with new vertices appearing on this collisions, and new subtractions are going to be required.

About the last equation, this is an identity between operators in the Heisenberg picture. Again, classically you expect that the left-hand side is identically zero as an operator equation. The fact that the current operator suffers from ordering problems in the case of chiral fermions, you are going to get a correction. What is interesting is that this correction is exact at one-loop computation. Usually equations like the first one you wrote will continue to get corrections indefinitely. You can think about this equations (already with the proper corrections) as equation of motions coming from varying the 1PI quantum effective action.

So, in your example of the chiral anomally, the 1PI quantum effective action at one-loop will have a term that explicitly breaks the chiral anomally and contribute with the equations of motion.

The devil is in the details. When you say that

...modulo formal issues involving multiplying operator-valued distributions

right after you wrote the equations of motion you ignore a very important point. The left-hand side of the equation

$$ (\partial^2 + m^2) \hat{\phi} = \hat{\phi}^3 $$

is not well defined if you does not regularize and renormalize it. Turns out that in other to do it exactly you need in advance the complete (quantum) solution of the model. One way to circumvent this problem is by organize your computations by perturbation theory of the non-linearity. Then at each order in your computation, for a given regularization, you are going to need to introduce a counter-term to match some renormalization conditions (and avoid nonsensical divergences). The anomaly are going to appear by the fact that there is no regularization, and no finite conter-term, able to preserve the anomalous symmetry. This imply that any attempt to make sense of the equation of motion you wrote are going to break the classical symmetry.

This conclusion does not depends on the perturbation theory. You may try to make sense of this equation by putting it on the lattice use computers to do non-perturbative computations. By putting the theory on the lattice you are going to regularize the theory first. The anomaly will appear as an impossibility to put this theory on the computer without breaking (explicitly) the symmetry.

A more abstract way to see what goes wrong. In free theory operators like $\phi^3$ need to be normal ordered. Once you introduce an interaction, the normal ordering are going to be modified, and you can calculate this modifications perturbativally. The calculation is obtained by looking at $\phi^3$ as a vertex of Feynman diagrams, and the normal ordering of free theory as self-contraction of the legs that come out of the vertex. In the interacted case there will be also collision of this legs, with new vertices appearing on this collisions.

The devil is in the details. When you say that

...modulo formal issues involving multiplying operator-valued distributions

right after you wrote the equations of motion you ignore a very important point. The right-hand side of the equation

$$ (\partial^2 + m^2) \hat{\phi} = \hat{\phi}^3 $$

is not well defined if you does not regularize and renormalize it. Turns out that in other to do it exactly you need in advance the complete (quantum) solution of the model. One way to circumvent this problem is by organize your computations by perturbation theory of the non-linearity. Then at each order in your computation, for a given regularization, you are going to need to introduce a counter-term to match some renormalization conditions (and avoid nonsensical divergences). The anomaly are going to appear by the fact that there is no regularization, and no finite conter-term, able to preserve the anomalous symmetry. This imply that any attempt to make sense of the equation of motion you wrote are going to break the classical symmetry.

This conclusion does not depends on the perturbation theory. You may try to make sense of this equation by putting it on the lattice use computers to do non-perturbative computations. By putting the theory on the lattice you are going to regularize the theory first. The anomaly will appear as an impossibility to put this theory on the computer without breaking (explicitly) the symmetry.

A more abstract way to see what goes wrong. In free theory operators like $\phi^3$ need to be normal ordered. Once you introduce an interaction, the normal ordering are going to be modified, and you can calculate this modifications perturbativally. The calculation is obtained by looking at $\phi^3$ as a vertex of Feynman diagrams, and the normal ordering of free theory as self-contraction of the legs that come out of the vertex. In the interacted case there will be also collision of this legs, with new vertices appearing on this collisions.

About the last equation, this is an identity between operators in the Heisenberg picture. Again, classically you expect that the left-hand side is identically zero as an operator equation. The fact that the current operator suffers from ordering problems in the case of chiral fermions, you are going to get a correction. What is interesting is that this correction is exact at one-loop computation. Usually equations like the first one you wrote will continue to get corrections indefinitely. You can think about this equations already corrected as coming from the 1PI quantum effective action.

So, in your example of the chiral anomally, the 1PI quantum effective action at one-loop will have a term that breaks the chiral anomally and contribute with the equations of motion.