Yes it is of this form because any Lie polynomial
"begins" by a Lyndon word, in particular, the least
monomial of $B(l)$ is $l$. Then
$$
B(m)=m+\sum_{m<u\atop u=m} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop v=n} \alpha^{(v)}v\qquad \mathbf{(LB)}
$$
then $[B(m),B(n)]$ has only words of length $m+n$ and its least word is $mn$.
Computationally

$$
B(m)B(n)=mn+[\sum_{m<v\atop v=n} \alpha^{(v)}mv+\sum_{m<u\atop u=m} \alpha^{(u)}un]=mn+[sb1]
$$
all the monomials within square brackets are of same length and strictly greater than $mn$

$$
B(n)B(m)=nm+[\sum_{m<v\atop v=n} \alpha^{(v)}vm+\sum_{m<u\atop u=m} \alpha^{(u)}nu]=nm+[sb2]
$$
all the monomials within square brackets are of same length and strictly greater than $nm$.

But, $mn<nm$ because $mn$ is Lyndon and then
$$
[B(m),B(n)]=mn+[sb1]nm[sb2]=mn+[sb3]
$$
where the square bracket $[sb3]$ is a linear combination of
monomials that are greater than $mn$ or $nm$
Hence all (monomials of $[sb3]$) are greater
than $mn$ which is the form you required.
One can say a little bit more From property $\mathbf{(LB)}$, and the fact that $[B(m),B(n)]$ is a multihomogeneous Lie polynomial, one has ($\underline{w}$ being the commutative image of $w$)
$$
[B(m),B(n)]=B(mn) + \sum_{mn<l\atop l\ \mathrm{Lyndon};\ \underline{l}=\underline{mn}} \gamma_{m,n}^{(l)}B(l)
$$
the coefficients $\gamma_{m,n}^{(l)}$ are the structure constants of the free Lie algebra w.r.t. the LyndonSirsov basis (they are integers, universal, i.e. characteristic free) and, up to my knowledge, their combinatorics is widely unknown.