Given a noncommutative nonunital ring $R$ and a subset $S\subseteq R$, I know that the left and right ideals generated by $S$ in $R$ have the forms$$RS:=\{\sum^n_{i=1} r_is_i:n\in\mathbb{N}\backslash\{0\},r_i\in R,s_i\in S \}$$and$$SR:=\{\sum^n_{i=1}s_ir_i:n\in\mathbb{N}\backslash\{0\},r_i\in R,s_i\in S \}$$respectively.
I also know that the two sided ideal generated by $S$ in $R$ has the form$$RSR:=\{\sum^n_{i=1}r_is_i\tilde{r}_i:n\in\mathbb{N}\backslash\{0\},r_i,\tilde{r}_i\in R,s_i\in S \}.$$
Now, my question is, how do the above ideals differ from a general ideal generated by $S$ in $R$?
For example, maybe $R=x\mathbb{C}\langle x,y\rangle$ is the free algebra over $\mathbb{C}$ whose basis consists of words in $x$ and $y$ beginning with $x$ on the left and whose multiplication is concatenation of words.
Take $S=x\mathbb{C}\langle x\rangle=x\mathbb{C}[x]$ to be the set of nonconstant polynomials over $\mathbb{C}$.
Then, clearly none of $RS$, $SR$ or $RSR$ contain $S$ as a subset as they would not contain the monomial $x$.
So, my guess is that a general ideal generated by $S$ in $R$ would look like$$\mathcal{I}\left(S\right)=RS+SR+RSR$$which is all the ways we can combine $R$ and $S$ as finite linear combinations under the ring operations.
Clearly, this works as a left, right and two-sided ideal since$$R\mathcal{I}\left(S\right)=RS+RSR\subseteq \mathcal{I}\left(S\right),$$$$\mathcal{I}\left(S\right)R=SR+RSR\subseteq \mathcal{I}\left(S\right)$$and$$R\mathcal{I}\left(S\right)R=RSR\subseteq \mathcal{I}\left(S\right).$$
Having said all that, I have seen some sources include the term $\mathbb{Z}S$ in the summand but surely this cannot be the case as $1\in\mathbb{Z}$ and so this would mean $1s=s\in \mathbb{Z}S$ for each $s\in S$ which would imply $S\subseteq \mathbb{Z}S\subseteq \mathbb{Z}S+ RS+SR+RSR$.
What am I missing here?