What's stopping me from choosing the nth Eilenberg Mac Lane space to be the following simplicial abelian group? Given an abelian group $X$, let $F_n(X)$ denote the simplicial abelian group defined as follows:
$F_n(X)_j=0$ for all $j<n$ and $F_n(X)_j=X$ for all $j\geq n$
with the appropriate zero and identity maps between them so the normalization (normalized moore complex) gives the appropriate homology (say, $d_i=0$ for every $i\geq 1$ (obviously for the parts higher than $n$) and $d_0=id_X$).  Then by the Hurewicz theorem, this simplicial abelian group seems like it should have homotopy concentrated in degree $n$ with $n$th component isomorphic to $X$ by computing the homology of the normalization.  
Since this thing is a Kan complex (because it's a simplicial abelian group), isn't this a representative for the Eilenberg-Mac Lane space $\kappa(X,n)$?
 A: 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.
A: The object you describe can't actually be made into a simplicial object because the boundary maps are incompatible with being able to define degeneracies.
Let's examine $n=0$.  Then you're going to need to define a degeneracy map $s_0: X \to X$ that satisfies the simplicial identities $d_0 s_0 = d_1 s_0 = id$.  However, substituting in your value for $d_1$, this says $id = 0$.
More generally, if you have a simplicial object which is zero in degrees less than $n$ and $X$ in degree $n$, then the degeneracies give rise in degree $m$ to - at least - one summand isomorphic to $X$ per surjection of ordered sets $\{0\ldots m\} \twoheadrightarrow \{0\ldots n\}$.
However, you can define a simplicial object which, in degree $m$, is
$$
\bigoplus_{\{0\ldots m\} \twoheadrightarrow \{0\ldots n\}} X
$$
with appropriate boundary maps, and this does give you an Eilenberg-Mac Lane space for $X$.  This is some kind of "direct sum of copies of $X$ indexed by the simplices of $S^n$".
