Consider the evaluation map $S^1 \times [S^1,K(Z,n)] \to K(Z,n)$. Since $S^1$ is a $K(Z,1)$, and $[S^1, K(Z,n)]=\Omega^1(K(Z,n))$ is a $K(Z,n-1)$, up to homotopy we get a map $K(Z,1)\times K(Z,n-1) \to K(Z,n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:

Consider the space of pointed maps $[K(Z,n),K(Z,m+n)]$. Then $\pi_k([K(Z,n),K(Z,m+n)])=0$ for $k>m$, and $=Z$ for $k=m$. To see this, note that (all maps are pointed) $$[S^k, [K(Z,n),K(Z,m+n)]] = [K(Z,n), \Omega^k(K(Z,m+n))]$$ $$ =[K(Z,n),K(Z,m+n-k)]= {\check H}^{m+n-k}(K(Z,n),Z) = 0 $$

if $k>m$, and $=Z$ if $k=m$.

Thus, we have a map $i: K(Z,m) \to [K(Z,n), K(Z,m+n)]$ by obstruction theory sending $\pi_m(K(Z,m))\to \pi_m([K(Z,n),K(Z,m+n)])$ isomorphically, which of course is equivalent to a map

$$ K(Z,m) \times K(Z,n) \to K(Z,m+n).$$

Maybe someone could explain to me if this gives the correct cohomology operation?

Consider the evaluation map $S^1 \times [S^1,K(Z,n)] \to K(Z,n)$. Since $S^1$ is a $K(Z,1)$, and $[S^1, K(Z,n)]=\Omega^1(K(Z,n))$ is a $K(Z,n-1)$, up to homotopy we get a map $K(Z,1)\times K(Z,n-1) \to K(Z,n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:

Consider the space of pointed maps $[K(A,n),K(A,m+n)]$. Then $\pi_k([K(A,n),K(A,m+n)])=0$ for $k>m$, and $=A$ for $k=m$. To see this, note that (all maps are pointed) $$[S^k, [K(A,n),K(A,m+n)]] = [K(A,n), \Omega^k(K(A,m+n))]$$ $$ =[K(A,n),K(A,m+n-k)]= {\check H}^{m+n-k}(K(A,n),A) = 0 $$

if $k>m$, and $=A$ if $k=m$ (by Hurewicz).

Thus, we have a map $i: K(A,m) \to [K(A,n), K(A,m+n)]$ by obstruction theory sending $\pi_m(K(A,m))\to \pi_m([K(A,n),K(A,m+n)])$ isomorphically, which of course is equivalent to a map (by evaluation)

$$ K(A,m) \times K(A,n) \to K(A,m+n).$$

Maybe someone could explain to me if this gives the correct cohomology operation?

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?

Consider the evaluation map $S^1 \times [S^1,K(Z,n)] \to K(Z,n)$. Since $S^1$ is a $K(Z,1)$, and $[S^1, K(Z,n)]=\Omega^1(K(Z,n))$ is a $K(Z,n-1)$, up to homotopy we get a map $K(Z,1)\times K(Z,n-1) \to K(Z,n)$. I'm not actually sure if this induces the product on cohomology. If it does, there is a natural generalization:

Consider the space of pointed maps $[K(Z,n),K(Z,m+n)]$. Then $\pi_k([K(Z,n),K(Z,m+n)])=0$ for $k>m$, and $=Z$ for $k=m$. To see this, note that (all maps are pointed) $$[S^k, [K(Z,n),K(Z,m+n)]] = [K(Z,n), \Omega^k(K(Z,m+n))]$$ $$ =[K(Z,n),K(Z,m+n-k)]= {\check H}^{m+n-k}(K(Z,n),Z) = 0 $$

if $k>m$, and $=Z$ if $k=m$. 

Thus, we have a map $i: K(Z,m) \to [K(Z,n), K(Z,m+n)]$ by obstruction theory sending $\pi_m(K(Z,m))\to \pi_m([K(Z,n),K(Z,m+n)])$ isomorphically, which of course is equivalent to a map

$$ K(Z,m) \times K(Z,n) \to K(Z,m+n).$$

Maybe someone could explain to me if this gives the correct cohomology operation?