This is an interesting question, to which the answer is **positive**.

Here is the proof. Of course, for the method to make sense, we must assume that the diagonal $D>0$. The notations below are borrowed from Section 12.3.2 of my book *Matrices* (2nd edition, Springer-Verlag, GTM **216**). Let $G=(D-E)^{-1}E^T$ the iteration matrix. One checks easily that $\ker A$ is the eigenspace of $G$, associated with the eigenvalue $\lambda=1$. The corresponding eigenspace for $G^T$ is $(D-E)^T\ker A$, fro which we find that $G$ has the invariant subspace $(D-E)^{-1}R(A)$. This can be verified directly with the help of the formula $G(D-E)^{-1}A=(D-E)^{-1}AG$. It turns out that $1$ is semi-simple: if $Gw=w+V$ and $Gv=0$, we obtain $v^T(D+A)v=0$, hence $v=0$. Therefore
$${\mathbb R}^n=\ker A\oplus (D-E)^{-1}R(A)$$
is a decomposition into $G$-invariant subspaces.

There remains to prove that for every vector $x^0$, the sequence $x^m:=G^mx^0$ is convergent. Let us decompose $x^m=y^m+z^m$, according to the invariant subspaces above. We have $y^m=y^0$, so this part is obviously convergent. There remains to prove that the spectral radius of the restriction $g$ of $G$ to $(D-E)^{-1}R(A)$ is smaller than $1$. This is proved exactly the same way as in Lemma 20 of the reference. We have to prove that if $x\in(D-E)^{-1}R(A)$, then $(Gx^T)A(Gx)\le x^TAx$, and equality implies $x=0$. With $y=(D-E)^{-1}Ax$, we have
$$(Gx^T)A(Gx)-x^TAx=-y^TDy\le0.$$
If the right-hand side vanishes, then $y=0$, which means $x\in\ker A$, hence $x=0$. Q.E.D