Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$, then by PBW theorem we know $U(\mathfrak{h})\hookrightarrow U(\mathfrak{g})$.
Let $\{x_i, y_i\}$ be an ordered basis of $\mathfrak{g}$ where $\{x_i\}$ is a basis of $\mathfrak{h}$. Then $\{1, x_{i_1}\cdots x_{i_k}y_{j_1}\cdots y_{j_m}\}$, where $i_1\le \cdots \le i_k$ and $j_1\le \cdots\le j_m$, forms a PBW basis of $U(\mathfrak{g})$.
Since $U(\mathfrak{h})$ is a subalgebra of $U(\mathfrak{g})$, we can define left or right action of $U(\mathfrak{h})$ on $U(\mathfrak{g})$ by left or right multiplication. And PBW tells us that this action makes $U(\mathfrak{g})$ a free left (or right) $U(\mathfrak{h})$-module. Let us call this action 1.
Let action 2 be induced by: $$x_{i_1}\cdots x_{i_k}\cdot (x_{j_1}\cdots x_{j_l}y_{m_1}\cdots y_{m_n}) = sort(x_{i_1}\cdots x_{i_k}\cdot x_{j_1}\cdots x_{j_l})y_{m_1}\cdots y_{m_n}$$ We can define the right action similarly, and it coincide with left action because we sort the basis elements by order. Also by PBW we can observe that action 2 makes $U(\mathfrak{g})$ a free left (or right) $U(\mathfrak{h})$-module, with the same free basis as was with action 1.
So does it follow that the two $U(\mathfrak{h})$-module structure on $U(\mathfrak{g})$ are isomorphic?
Your action 2 is not even well-defined if $\mathfrak{h}$ is not commutative:
Let $i<j$ such that $[x_i,x_j]\neq0$. Then $$(x_ix_j)\cdot1=x_ix_j=(x_jx_i)\cdot1,$$ hence: $$0\neq[x_i,x_j]=[x_i,x_j]\cdot1=(x_ix_j-x_jx_i)\cdot1=0,$$ which is impossible.