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 $$ B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v $$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$.

Computationally

1. $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
2. $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
3. 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.  

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

1. $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
2. $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
3. 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 Lyndon-Sirsov basis (they are integers, universal, i.e. characteristic free) and, up to my knowledge, their combinatorics is widely unknown.

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 $$ B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v $$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$.

Computationally

1. $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
2. $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
3. 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 greater than $mn$ which is the form you required.

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 $$ B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v $$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$.

Computationally

1. $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
2. $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
3. 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.