You can find such a map, but it goes the other way: $\Gamma(Y)\to \Omega X$. It is always wise to keep one's right adjoints on the right hand side.t
But first, let me note that every simplicial abelian group $Z$ is equivalent to $\prod_{k\ge0} K(\pi_kZ,k)$ (this is just a restatement of the classical fact that every chain complex of $\mathbb{Z}$-modules is quasi-isomorphic to its homology), so just the observation about homotopy groups suffices to conclude that $\Gamma(Y)$ and $\Omega X$ are equivalent.
Now, for the construction of the map, note that there is a canonical diagram
$$\require{AMScd}
\begin{CD}
Y @>>> 0\\
@VVV @VVV\\
0 @>>> N(X)
\end{CD}
$$
equipped with a chain homotopy of the zero map $Y\to N(X)$ with itself (the latter is just the canonical isomorphism $Y[1]\to N(X)$).
Since Dold-Kan sends chain homotopies to simplicial homotopies, after applying $\Gamma$ we get a homotopy commutative square
$$\require{AMScd}
\begin{CD}
\Gamma(Y) @>>> *\\
@VVV @VVV\\
* @>>> X
\end{CD}
$$
which again induces the required map from $\Gamma(Y)$ to the homotopy pullback of the rest of the diagram, that is $\Omega X$.
There is a more concrete (although messier) way of obtaining the above map, and it is via constructing the adjoint map $\Sigma \Gamma(Y)\to X$. This is because you can get the suspension via the Kan suspension functor ([1], III.5)
$$(\Sigma K)_n = K_n \vee K_{n-1}\vee \cdots \vee K_0$$
so to get map $\Sigma \Gamma(Y)\to X$ you just need to give maps
$$(\Gamma(Y))_l=\bigoplus_{[l]\twoheadrightarrow [i]} Y_i = \bigoplus_{[l]\twoheadrightarrow [i]} (NX)_{i+1}\to X_n =\bigoplus_{[n]\twoheadrightarrow [i]} (NX)_{i}$$
for every $l\le n$ satisfying some compatibilities. Spoiler alert: these will just be the canonical inclusions.
In fact what we are doing is just identifying $X$ with $\bar W \Gamma(Y)$, where $\bar W$ is the Eilenberg-MacLane classifying space for a simplicial group ([1] Remark III.5.6).
[1] Goerss, Jardine Simplcial Homotopy Theory, 2006
