Let $\mathfrak{g}$ have the basis $x$, $(y_n)_{n\ge 0}$ with $[x,y_n]=y_{n+1}$ and $[y_n,y_m]=0$.
The ideals are $0$, the ideal $I_n$ with basis $(y_k:k\ge n)$, and all subspaces containing the derived subalgebra $I_1$.
Is this Lie algebra Noetherian (satisfy the maximal condition for ideals)?
Could I obtain a semi-prime ideal such that $r(P) \not = P$?
A Lie algebra $\mathfrak{g}$ satisfy the maximal condition for ideals, if for each , ascending chain $H_{1} \subseteq H_{2} \subseteq \ldots $ an index $m$ exists such that $H_{i}=H_{k}$ if $m<i$, $m<k .$
We say in short: $ \mathfrak{g} \in{\rm Max}-\triangleleft$ or $\mathfrak{g}$ is Noetherian Lie algebra.
There are many more ideals than you're claiming, still it's noetherian.
Indeed, let $V$ be the hyperplane with basis $(y_n)$, and $T$ the operator $y_n\mapsto y_{n+1}$, and $K$ the ground field. Then a subspace of $V$ is an ideal iff it is a $T$-invariant subspace of $V$. Make $V$ a $K[x]$-module with $x$ acting as $T$. Then $V$ is generated by $y_0$ and hence, being infinite-dimensional over $K$, is a free $K[x]$-module of rank 1. Hence it's a noetherian module: precisely its submodules are $\{0\}$ and for each element of the form $y=y_n+\sum_{i=n+1}^Na_jy_j$, the submodule generated by $y$ (this just reflects that $K[x]$ is a PID and that every nonzero ideal is generated by a unique monic polynomial). Only for $y=y_n$ this corresponds to one of the ideals you list. For instance, the ideal generated by $y_0-y_1$ contains no $y_n$.
The remaining ideals, i.e., those not contained in $V$, contain $y_1$ and hence have codimension $\le 2$. Hence, every nonzero ideal has finite codimension, so $\mathfrak{g}$ is noetherian.