Universal enveloping algebra of a Poisson algebra For a Lie algebra, $\mathfrak{g}$, one has an equivalence of categories between Mod($\mathfrak{g}$) and Mod($U(\mathfrak{g})$), where $U(\mathfrak{g})$ is the universal enveloping algebra of $\mathfrak{g}$. 
Let $P=(A,\{,\})$ be a Poisson algebra over $A=k[x_1,...,x_n]$.
An $A$-module $P$ is a Poisson module if there is a bilinear product $\{,\}_M:A \times M \rightarrow M$ such that the following hold for all $a,b \in A$ and all $m \in M$:


*

*$\{\{a,b\},m\}_M = \{a,\{b,m\}_M\}_M - \{b,\{a,m\}_M\}_M$;

*$\{a,bm\}_M = \{a,b\}_M + b\{a,m\}_M$;

*$\{ab,m\}_M = a\{b,m\}_M + b\{a,m\}_M$.


One can construct the enveloping algebra of $P$ in much the same way as one constructs $U(\mathfrak{g})$. In particular, 
$U(P) = k\langle x_1,...,x_n \mid x_ix_j-x_jx_i - \{x_i,x_j\} \text{ for all } 1 \leq i,j \leq n\rangle$,
$k$ a field.
Does the same equivalence exist between $U(P)$-modules and Poisson $P$-modules?
Here is the example I have in mind. Let $A=k[x,y,z]$ and define a Poisson bracket on $A$ by $\{x,y\}=z^2$, $\{y,z\}=x^2$, $\{z,x\}=y^2$. Then the universal enveloping algebra for $P$ should be 
$k\langle x,y,z \mid xy-yx=z^2, yz-zy=x^2, zx-xz=y^2 \rangle$.

The answer to your question is negative. 
For any Poisson algebra $P$, there does exist an associative algebra, which is denoted by $P^e$, such that the module category over $P^e$ is equivalent to the category of Poisson modules over $P$. See http://arxiv.org/pdf/1102.0366v1.pdf for the details.
However, $P^e$ is not constructed the way you mentioned. In fact, one should think $P^e$ as the analog of $A\otimes A^{op}$ of an associative algebra $A$. So in some sense, the size of $P^e$ is "doubled" from $P$.
As for your example, $P^e$ is a non-commutative algebra of six generators of GK-dimension 6. 
