Consider the evaluation map $S^1 \times [S^1,K(\mathbb{Z},n)] \to K(\mathbb{Z},n)$. Since $S^1$ is a $K(\mathbb{Z},1)$, and $[S^1, K(\mathbb{Z},n)]$ is a $K(\mathbb{Z},n-1)$, we get a map $K(\mathbb{Z},1)\times K(\mathbb{Z},n-1) \to K(\mathbb{Z},n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:
There's a canonical homotopy type of (pointed) map $S^n \to K(\mathbb{Z},n)$ by taking a generator $1\in \mathbb{Z}=\pi_n(K(\mathbb{Z},n))$. This composition induces a map $[K(\mathbb{Z},n),K(\mathbb{Z},m+n)]\to [S^n,K(\mathbb{Z},m+n)]\simeq K(\mathbb{Z},m)$. Thus, again we get an evaluation map $$K(\mathbb{Z},n)\times K(\mathbb{Z},m) \simeq K(\mathbb{Z},n)\times [S^n,K(\mathbb{Z},m+n)] \to K(\mathbb{Z},m)\times [K(\mathbb{Z},m),K(\mathbb{Z},m+n)] \to K(\mathbb{Z},m+n).$$
Maybe someone could explain to me if this gives the correct cohomology operation?