Let me put Tyler Lawson's answer into a general context. You can define the chain complex $K(A,n)$ by letting
$$K(A,n)_i = \begin{cases} A & i = n \\ 0 & i \neq n \end{cases}$$
and with the zero differential. This is of course a well-defined chain complex: $0^2 = 0$ (you don't run into issues as for simplicial abelian groups). I suspect that intuitively this is what you wanted to do.
The Dold–Kan correspondence says that the category of chain complexes is equivalent to the category of simplicial abelian group:
$$N : \mathsf{Ab}^{\Delta^{op}} \leftrightarrows \mathsf{Ch}_{\ge 0} : \Gamma.$$
Here $N$ is the functor of normalized chains and $\Gamma$ can be described explicitly. It turns out that if you apply $\Gamma$ to the chain complex defined above, you exactly get the simplicial abelian group described in Tyler Lawson's answer. A possible reference is the book Simplicial Homotopy Theory of Goerss and Jardine, Chapter III, Section 2. Explicitly, for a chain complex $C_*$, the simplicial abelian group $\Gamma(C)$ is given by:
$$\Gamma(C)_n = \bigoplus_{[n] \twoheadrightarrow [k]} C_k$$
where the direct sum ranges over surjection from $[n] = \{ 0, \dots, n \}$ to $[k]$. The simplicial structure maps are defined through some kind of yoga with epi-mono factorizations. So for $C_* = K(A,n)_*$ you exactly get the simplicial abelian group of Tyler Lawson's answer.
It moreover follows from the construction (this is also explained in the book) of the equivalence $N \dashv \Gamma$ that the homology groups $H_*(C)$ of a chain complex $C_*$ are isomorphic to the homotopy groups $\pi_*(\Gamma(C_*))$; but of course the homology of the chain complex $K(A,n)$ is simply $A$ concentrated in degree $n$, thus $\Gamma(K(A,n))$ is an Eilenberg–MacLane simplicial abelian group.