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?