Relationship between $BU(1)$ and $BU$

Question

Morally, there should be a kind of "splitting principle" relationship between $\mathbb{C}P^\infty \simeq BU(1)$ (which classifies line bundles) and $BU$ (which classifies vector bundles). For example, we have something like $$H_*(BU) \cong \operatorname{Sym}(\tilde{H}_*(BU(1))).$$

My question is: how is this be realized topologically? That is, how should I fill in the following blank?

$BU$ is the free _______ generated by $\mathbb{C}P^\infty$.

Here are some thoughts on this matter.

1. Consider the free loop space (i.e., $\mathbb{A}_\infty$-space) $\Omega \Sigma \mathbb{C}P^\infty$ on $\mathbb{C}P^\infty$. Since $\mathbb{C}P^\infty$ is connected, there is natural weak equivalence betweeen $\Omega \Sigma \mathbb{C}P^\infty$ and the James construction $J\mathbb{C}P^\infty$, where $J\mathbb{C}P^\infty \simeq \bigvee_{n \geq 0} BU(1)^{\times n}/\sim$. Squinting at this, this looks a bit like $\bigsqcup_{n \geq 0} BU(n)$, whose group completion, by Bott periodicity, is the space $BU \times \mathbb{Z}$. In fact, the maps $BU(1)^{\times n} \to BU(n)$ factor though $\Sigma_n$-orbits, so to pass from $J\mathbb{C}P^\infty$ to $\bigsqcup_{n \geq 0} BU(n)$ I ought to impose some kind of commutativity.

2. As Hopf algebras, $H_*(BU)$ can be identified with the Hopf algebra $\mathrm{Symm}$ of symmetric functions, which is self-dual, i.e., $H_*(BU) \cong H^*(BU)$. The evident map $\Omega \Sigma \mathbb{C}P^\infty \to BU$ induces maps $H_*(\Omega \Sigma \mathbb{C}P^\infty) \to H_*(BU)$ - which is identified with the surjection $\mathrm{NSymm} \to \mathrm{Symm}$ - and $H^*(BU) \to H^*(\Omega \Sigma \mathbb{C}P^\infty)$ - which is identified with the inclusion $\mathrm{Symm} \to \mathrm{QSymm}$. Here $\mathrm{NSymm}$ and $\mathrm{QSymm}$ are the Hopf algebras of non-symmetric functions and quasi-symmetric functions respectively. Again, this seems to suggest that $\Omega \Sigma \mathbb{C}P^\infty$ is too big to be $BU$ because it isn't commutative enough.

3. So consider instead the free infinite loop space (i.e., $\mathbb{E}_\infty$-space) $Q\mathbb{C}P^\infty = \Omega^\infty \Sigma^\infty \mathbb{C}P^\infty$. This is still a little too large to be $BU$, but in fact contains $BU$ as a retract of loop spaces. Moreover, $BU$ and $Q\mathbb{C}P^\infty$ become homotopy equivalent after rationalization. So we're getting closer but still not quite there.

Am I overlooking something obvious?