Structure of ideals in the product of two rings
If you have an identity, multiplying by $(1_R,0)$ and $(0,1_S)$ will help. If you don't, then the result need not be true. E.g. if you put the zero product on the abelian group $\mathbb Z_2\oplus \mathbb Z_2$, then the ideal ($=$subgroup) generated by $(\overline 1,\overline 1)$ is a counterexample.
Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $R\times S$. Define $I_R = \left\{r \in R: \, \exists s \in S \, \, \, \text{s.t.} \, \, (r,s) \in I \right\}$ and similarly $I_S = \left\{s \in S: \, \exists r \in R \, \, \, \text{s.t.} \, \, (r,s) \in I \right\}$. Then $I_S, I_R$ are ideals of $R,S$ and $I \subset I_R \times I_S$. Now take $r \in I_R$ and $s \in I_S$. We will show that $(r,s) \in I$. By definition there is some $s' \in S$ such that $(r,s') \in I$. Similarly there is some $r' \in R$ such that $(r',s) \in I$. Since $I$ is a two-sided ideal, we have that $(1_R,s)(r,s') = (r,ss') \in I$. Similarly, $(r',s)(1_R,s') = (r',ss') \in I$. Hence $(r-r',0) \in I$ and so $(r,s) \in I$.
Remark: As we see, all we need is that only one of $R,S$ has an identity element.
Lemma. Let $(A_i)_{i=1,\ldots,n}$ be a (finite) family of rings with $1$. Then ideals of $A = \prod_{i=1}^n A_i$ are of the form $I_1 \times \ldots\times I_n$, where $I_i$ is an ideal of $A_i$ for each $i\in\{1,\ldots,n\}$.
Proof. By induction on $n$, the cases $n\in\{0,1\}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $A\times B$ of two rings with $1$, and denote by $p : A\times B\to A$ and $q : A\times B\to B$ the two canonical projections. Then $I = p(K)$ (resp. $J = q(K)$) is an ideal of $A$ (resp. of $B$.), because the image of an ideal under a surjective ring homomorphism is an ideal. Obviously $K\subseteq I\times J$. To show the inverse inclusion, let $(a,b)\in I\times J$. Then $(a,b') \in K$ and $(a',b)\in K$ for some $(a',b')\in A\times B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) \in K$. $\square$