Given a simplicial ring $A_\bullet$, it is standard that $\pi_0(A)$ is a ring and $\pi_n(A)$ is a module over it, so $A_\bullet$ is weakly contractible iff $\pi_0(A)=0$. Also, $\pi_0(A)$ is just the cokernel of $d_0-d_1:A_1\to A_0$. We can take any chain of face maps $A_n\to A_0$ and compose with the projection to $\pi_0(A)$, and this will be independent of which chain we chose. These maps $A_n\to\pi_0(A)$ give a surjective map from $A_\bullet$ to the constant simplicial ring $\pi_0(A)$. (This is just the simplest piece of the Postnikov tower.) If $\pi_0(A)\neq 0$ we can then compose with a surjective map to a constant simplicial field.
Note also that all simplicial fields are constant. This is just because the composite $A_0\xrightarrow{s_0^n}A_n\xrightarrow{d_0^n}A_0$ is the identity, so $d_0^n$ is surjective, but all field homomorphisms are injective, so $d_0^n$ is an isomorphism.
I think it also works out that a simplicial ring is contractible as a simplicial group iff it is weakly contractible (because you can find an element $x\in A_1$ with $d_0(x)-d_1(x)=1$, and then shuffle-multiplication by $x$ should give a contraction). It would be stronger to say that there is a contraction by ring maps. I don't immediately see the full picture about that.