The cohomology ring for $BB\mathbb{Z}$ is $\mathbb{Z}[[x]]$, where $x$ lies in degree $2$. My question is about $\mathbb{Z}[[x_1, \dots, x_n]]$. I was trying to find a topological group (ideally an internal group in CW-complexes) such that the cohomology ring of $BG$ is $\mathbb{Z}[[x_1, ..., x_n]]$, where $x_1, \dots, x_n$ lie in some degrees (but I was aiming for degree $2$).
$BU(n)$ classifies $n$-dimensional vector bundles, which suggests to me that the cohomology ring of $BU(n)$ could resemble $\mathbb{Z}[[x_1, \dots, x_n]]$, or also $\mathbb{Z}[[x_1, \dots, x_n]]^{S_n}$.
Where can I find an account of the cohomology ring of $BU(n)$ (if the above suggestion is true), or if that's not the case, then how might I obtain a cohomology ring of $\mathbb{Z}[[x_1, \dots, x_n]]$?
As mentioned in the comments, when $X$ has infinitely many non-zero cohomology groups, there are two non-equivalent definitions for the cohomology ring of $X$: the direct sum or the direct product of all the cohomology groups. You are using the latter (many other sources use the former, so make sure to check conventions).
Since $S^1 = U(1)$ is a model for $B\mathbb{Z}$, and $\mathbb{CP}^{\infty}$ is a model for $BS^1 = BU(1)$, we see that $\mathbb{CP}^{\infty}$ is a model for $BB\mathbb{Z}$. With your convention, it follows that $H^*(BB\mathbb{Z}; \mathbb{Z}) \cong H^*(\mathbb{CP}^{\infty}; \mathbb{Z}) \cong \mathbb{Z}[[x_1]]$ where $x_1$ has degree $2$. In $H^*(BU(1); \mathbb{Z}_2)$, the generator $x_1$ can be identified with the first Chern class $c_1$.
It follows from the Kunneth Theorem, and the fact that $H^*(\mathbb{CP}^{\infty} ; \mathbb{Z})$ is torsion-free, that $H^*((\mathbb{CP}^{\infty})^n; \mathbb{Z}) \cong \mathbb{Z}[[x_1, x_2, \dots, x_n]]$ where $\deg x_i = 2$. Since $BG_1\times BG_2 = B(G_1\times G_2)$, we see that $(\mathbb{CP}^{\infty})^n$ is a model for $(BU(1))^n = B(U(1)^n) = B((S^1)^n) = BT^n$.
The inclusion $T^n \to U(n)$ of a maximal torus induces a map $BT^n \to BU(n)$, and the induced map on cohomology is injective (this is effectively the splitting principle). One can then show that $H^*(BU(n); \mathbb{Z}) \cong \mathbb{Z}[[x_1, x_2, \dots, x_n]]^{S_n} \cong \mathbb{Z}[[c_1, c_2, \dots, c_n]]$ where $\deg c_i = 2i$, see here for example.
In the comments you asked about the cohomology ring of the space of unitary operators on a Hilbert space. If the Hilbert space is infinite-dimensional, then the space of unitary operators is contractible (with the norm topology or the strong operator topology), see here, so the cohomology is trivial. In the finite-dimensional case, the space of unitary operators is $U(n)$ and $H^*(U(n); \mathbb{Z}) \cong \Lambda_{\mathbb{Z}}[x_1, x_3, \dots, x_{2n-1}]$ where $\deg x_i = i$, see Corollary 4D.3 of Hatcher's Algberaic Topology.