This is some confusion about standard facts.
The cohomology ring of $\mathbb{CP}^\infty$ is a polynomial ring in one variable. However, it is also well-known that for generalized cohomology theories, if the inverse system $h^t(X^n)$ satisfies the Mittag-Leffler condition, we have $$ h^*(\varinjlim X^n) \cong \varprojlim h^*(X^n) $$ If we think of infinite projective space as the colimit of $\mathbb{CP}^n$, can we then conclude: $$ H^*(\mathbb{CP}^\infty; \mathbb{Z}) \cong \varprojlim H^*(\mathbb{CP}^n; \mathbb{Z}) \cong \varprojlim \mathbb{Z}[x] / (x^{n+1}) \cong \mathbb{Z}[[x]] $$
Your argument is fine. Since $\deg x =2 $, the graded rings $\mathbb{Z}[[x]]$ and $\mathbb{Z}[x]$ coincide. (Technically, the issue here is that the cohomology graded ring is really a series of abelian groups $H^p(X)$ and nice maps $H^p(X) \otimes H^q(X) \to H^{p+q}(X)$. To make this object into an ordinary ring, we can take either $\bigoplus H^*(X)$ or $\prod H^*(X)$. In finite dimensions, they coincide; here, we really want the latter for exactly the reason you describe.)