Your proof strategy via (1) and (2) is impossible. If $PA\cup\Sigma$ proves that Goodstein's theorem is false, then the proof will have finite length, and so there will be some finite $\Sigma_0\subset\Sigma$ such that $PA\cup\Sigma_0$ proves that Goodstein's theorem is false. This would imply by (1) that Goodstein's theorem is false in the standard model. But Goodstein's theorem holds in the standard model, as Goodstein proved.
A second point is that you may find that there are no specific "natural" models of PA at all other than the standard model. For example, Tennenbaum proved that there are no computable nonstandard models of PA; that is, one cannot exhibit a nonstandard model of PA so explicitly that the addition and multiplication of the model are computable functions. (See this related MO question.) But I do not rule out that there could be natural nonstandard models in other senses.