Let $M=A+iB$ where $A$ and $B$ are real matrices. The decomposition $M=ER$ is possible iff $$ \operatorname{rank}M=\operatorname{rank}\pmatrix{A\\ B}.\tag{1} $$ In the above, whether or not $M$ is singular is irrelevant (although there is no need to ask your question when $M$ is nonsingular, because there is a trivial solution $(E,R)=(M,I)$). Also, if you want the decomposition $M=RE$ instead of $M=ER$, condition $(1)$ should be modified to $\operatorname{rank}M=\operatorname{rank}\pmatrix{A&B}$.
Example. Consider $M=\pmatrix{1&i\\ 0&0}$. This is a rank-one matrix. If it admits the decomposition $ER$, then $R$ must also be rank-one. Therefore $R=uv^T$ for some nonzero real vectors $u$ and $v$. In turn, the row space of $M=Euv^T$ is the span of $v^T$. But this is impossible because the first row of $M$, namely $\pmatrix{1&i}$, is not a scalar multiple of any real vector. In this example one can verify that $$ \operatorname{rank}M=1<\operatorname{rank}\pmatrix{A\\ B}=\operatorname{rank}\pmatrix{1&0\\ 0&0\\ 0&1\\ 0&0}=2. $$
Proof of the necessity and sufficiency of $(1)$. Suppose $M=ER$. Let $E=X+iY$ where $X$ and $Y$ are real. By comparing both sides of the equality $M=ER$, we get $A=XR,\,B=YR$. Hence we obtain $(1)$ by the sandwich principle: \begin{align*} \operatorname{rank}M =\operatorname{rank}\left[\pmatrix{I&iI}\pmatrix{A\\ B}\right] \le\operatorname{rank}\pmatrix{A\\ B} =\operatorname{rank}\pmatrix{XR\\ YR} \le\operatorname{rank}R =\operatorname{rank}M. \end{align*}
Conversely, suppose $(1)$ is satisfied. Let $r=\operatorname{rank}(M)$. Then $s:=n-r=\dim\ker\pmatrix{A\\ B}$. Hence there exists a real invertible matrix $Q$ such that both $AQ$ and $BQ$ are in the form of $$ \pmatrix{\ast&0_{r\times s}\\ \ast&0_{s\times s}} $$ and the polynomial matrix $(A+xB)Q$ (where $x$ be an indeterminate) also assumes the same form. Since $\operatorname{rank}(A+iB)=\operatorname{rank}(M)=r$, in the block column marked by two asterisks above, some $r$-rowed minor of $(A+iB)Q$ is nonzero, and the corresponding minor in $(A+xB)Q$ is a nonzero polynomial. Consequently, this minor is nonzero for $x=i$ and for $x=$ some nonzero $t\in\mathbb R$. Hence there exists a real invertible matrix $P$ such that $$ P(A+tB)Q=\pmatrix{\ast&0_{r\times s}\\ 0_{s\times r}&0_{s\times s}} $$ where the subblock marked by an asterisk is invertible. But that means $PAQ$ and $PBQ$ are necessarily of the forms $$ PAQ=\pmatrix{A'&0\\ -tC'&0}\quad\text{and}\quad PBQ=\pmatrix{B'&0\\ C'&0} $$ for some $A',B'\in M_r(\mathbb R)$, $C\in M_{s,r}(\mathbb R)$ such that both $A'+tB'$ and $A'+iB'$ are invertible. Now let $$ PXQ=\pmatrix{A'&0\\ -tC'&I_s},\quad PYQ=\pmatrix{B'&0\\ C'&0},\quad Q^{-1}RQ=\pmatrix{I_r\\ &0}. $$ Then $X+iY=P^{-1}\pmatrix{A'+iB'&0\\ (i-t)C'&I_s}Q^{-1}$ is invertible, $XR=A,\,YR=B$. Hence $M=ER$.
