As Andrés points out in the comments, your question seems to be resolved by the observation that not every model of PA can arise as the $\mathbb{N}$ of a model of ZFC.

Meanwhile, one can attempt to understand more deeply exactly which models of PA do arise as the $\mathbb{N}$ of a model of set theory. Let us say that a model $N\models\text{PA}$ is *a* standard model of arithmetic (as opposed to *the* standard model of arithmetic), or alternatively is a ZFC-standard model of arithmetic, if $N=\mathbb{N}^M$ for some $M\models\text{ZFC}$.

Thus, a model of arithmetic is a standard model of arithmetic, if it is the standard model of arithmetic from the perspective of some model of set theory. These are the models of PA that seem to be at the heart of your question.

We can characterize exactly what these models are as follows.

**Theorem.** (Ali Enayat) The following are equivalent, for a countable nonstandard model $N\models\text{PA}$.

$N$ arises as $\mathbb{N}^M$ for some model $M\models\text{ZFC}$.

$N$ is computably saturated and satisfies all the arithmetic consequences of ZFC.

In statement 2, what we mean is that $N\models\varphi$, whenever $\varphi$ is an arithmetic statement that is provable from ZFC. For example, this includes the case where $\varphi$ is Goodstein's theorem, and this is exactly what is going on in your question. Obviously, any model of PA arising as the $\mathbb{N}$ of a model of ZFC set theory must satisfy all the arithmetic consequences of ZFC, and this is essentially how you are reasoning in statement 3 of the question. The interesting part is that this is also sufficient, when combined with computable saturation.

You can find a proof of the theorem in my paper: J. D. Hamkins and R. Yang, Satisfaction is not absolute, to appear in the Review of Symbolic Logic, pp. 1-34. (arxiv:1312.0670) (see proposition 3).