Idempotent semigroups in general are not residually finite. There are some examples in Golubov, È. A.; Sapir, M. V. Varieties of finitely approximable semigroups. Izv. Vyssh. Uchebn. Zaved. Mat. 1982, no. 11, 21–29. If you want an easy example, I can post it here.
Here is an example. Let $S=L\cup R\cup Z\cup \{0\}$ where $L=\{l_1,l_2,...\},R=\{r_1,r_2,...\}$ are infinite sets, $Z=\{x,y\}$. Define a product by $l_i^2=l_i, r_i^2=r_i, x^2=x,y^2=y$, $l_ir_j=x$ if $i=j$, $l_ir_j=y$ if $i\ne j$, $l_ix=x, yr_i=y$, all other products are 0 (in particular, $xy=0$ and $0$ plays the role of zero. The associativity is easy to check. So $S$ is an idempotent semigroup. Suppose that $S$ is residually finite and there exists a homomorphism $\phi$ onto a finite semigroup separating $x$ from $y$. Note that there exist $i\ne j$ such that $\phi(l_i)=\phi(l_j)$. Then $\phi(x)=\phi(l_ir_i)=\phi(l_jr_i)=\phi(y)$, a contradiction.