The PBW filtration is multiplicative and has commutative associated graded

Concatenating a word of length at most $m$ with one of length at most $n$ gives length at most $m+n$. Taking spans and quotient images proves $F_m{F}_n\subseteq F_{m+n}$.

For a generator $x$ and a word $b_1\cdots b_s$, repeated use of $[x,ab]=[x,a]b+a[x,b]$ gives $[x,b_1\cdots b_s]={\sum }_j{b}_1\cdots [x,b_j]\cdots b_s$. By [L2], every $[x,b_j]$ is again the image of one element of $\mathfrak{g}$, so this commutator lies in $F_s$.

For words $a=a^{\mathrm{\prime }}x$ of length $r>0$ and $b$ of length $s>0$, the identity $[a^{\mathrm{\prime }}x,b]=a^{\mathrm{\prime }}[x,b]+[a^{\mathrm{\prime }},b]x$, together with step 1.2 and induction on $r$, puts both terms in $F_{r+s-1}$. The scalar boundary cases commute, and bilinearity therefore gives $[F_m,F_n]\subseteq F_{m+n-1}$.

If $a\in F_m$ and $b\in F_n$, step 2.1 says that $ab$ and $ba$ have the same class in $F_{m+n}\mathrm{/}{F}_{m+n-1}$. By [L3], all homogeneous elements of $\mathrm{gr}U(\mathfrak{g})$ commute, hence the whole associated graded algebra is commutative.