The universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ is defined to be $$ \dfrac{\mathbb{C}\oplus\mathfrak{g}\oplus ( \mathfrak{g}\otimes \mathfrak{g})\oplus (\mathfrak{g}\otimes \mathfrak{g}\otimes \mathfrak{g})\oplus\ldots}{\langle a\otimes b-b\otimes a -[a,b]\rangle}, $$which means it is an associative tensor algebra generated by the vector space $\mathfrak{g}$ mod out all elements of the form $ a\otimes b-b\otimes a -[a,b]$, where $a,b\in \mathfrak{g}$.
Suppose $\mathfrak{g}=\mathbb{C}[e,p,q]/\langle [p,q]=e,[p,e]=0,[q,e]=0 \rangle$. Then we have a (ring) isomorphism
$$ U(\mathfrak{g})/\langle e-1\rangle \cong \mathbb{C}[x, \partial/\partial x] $$ where we can define the map to be $U(\mathfrak{g})\stackrel{\phi}{\rightarrow}\mathbb{C}[x,\partial/\partial x]$ by sending $$ p \mapsto \partial/\partial x \mbox{ and } q \mapsto x. $$ One can check that since $$ \phi([p,q])=[\partial/\partial x,x]= 1-x\dfrac{\partial}{\partial x}, $$ we take $e=x\dfrac{\partial}{\partial x}$, which gives us the isomorphism $U(\mathfrak{g})/\langle 1-e\rangle \cong \mathbb{C}[x,\partial/\partial x]$.
Now what should $\mathfrak{g}$ be so that $U(\mathfrak{g})/I\cong \mathbb{C}[x_1,x_2,\ldots, x_n,\partial /\partial x_1,\partial /\partial x_2, \ldots, \partial /\partial x_n]$?
I could be wrong but I'm guessing that if we take $\mathfrak{g}$ to be the algebra $\mathbb{C}[p_i,q_i,e_i]/I$ where $I$ is generated by brackets of the form $$ [p_i,q_j]= \left\{ \begin{aligned} e_i &\mbox{ if } i=j\\ -q_j p_i &\mbox{ if } i\not= j \\ \end{aligned} \right. $$
$$ [p_i,e_j]= \left\{ \begin{aligned} 0 &\mbox{ if } i=j\\ - p_i &\mbox{ if } i\not= j \\ \end{aligned} \right. $$
$$ [q_i,e_j]= \left\{ \begin{aligned} 0 &\mbox{ if } i=j\\ q_i &\mbox{ if } i\not= j, \\ \end{aligned} \right. $$ then $U(\mathfrak{g})/\langle e_i-1\rangle\cong \mathbb{C}[x_1,x_2,\ldots, x_n,\partial /\partial x_1,\partial /\partial x_2, \ldots, \partial /\partial x_n]$ where we take the map to be $$ p+\langle e_i-1\rangle \mapsto \partial/\partial x_i $$ and $$ q+\langle e_i-1\rangle \mapsto x_i. $$
Also, since the order of multiplication matters, i.e.,
$$ (\overbrace{\partial/\partial x_1 + \partial/\partial x_2}^{\deg -1?})(\overbrace{x_1}^{\deg 1}) = 1+0=\overbrace{1}^{\deg 0} $$ while $$ x_1(\partial/\partial x_1 + \partial/\partial x_2) = x_1 \partial /\partial x_1 + x_1\partial/\partial x_2, $$ how should one think about multiplication in the skew polynomial algebra?
Finally, I just noticed that $x_i\dfrac{\partial}{\partial x_i}$ form mutually orthogonal idempotent elements in $\mathbb{C}[x_i,\partial/\partial x_i]$. That is,
$$ \left(\sum_i x_i \dfrac{\partial}{\partial x_i}\right)^2 = \left(\sum_i x_i \dfrac{\partial}{\partial x_i}\right).$$ I am guessing that these are the only non-scalar idempotent elements in the algebra.
Do these idempotent elements have a geometric interpretation?