To my taste, it seems more natural to let $A$ and $B$ play symmetric role, by asking whether there exists non-trivial factors $s_jA+t_jB$ such that $$A\left(\prod_{j=1}^p(s_jA+t_jB)\right)B=B\left(\prod_{j=1}^p(s_jA+t_jB)\right)A.$$
If you pose the question in an algebraically closed field $k$ (say, $k=\mathbb C$), then the answer is yes for the following reason:
There exist $n$ 2^n-1$non-zero factors$s_jA+t_jB$such that$\prod_{j=1}^n(s_jA+t_jB)=0$. The proof is by induction over the rank of products$\prod_{j=1}^p(s_jA+t_jB)=0$. Suppose that exists such a product$\Pi$, with rank$r\ge1$. Let us write $$\Pi=\sum_{j=1}^rx_ja_j^T.$$ Then $$\Pi M\Pi=\sum_{i,j=1}^r(a_i^TMx_j)x_ia_j^T.$$ The rank of$\Pi M\Pi$will be less than or equal to$r-1$if$\det(a_i^TMx_j)_{1\le i,j\le r}=0$. When$M=sA+tB$, this writes$H(s,t)=0$where$H$is a homogeneous polynomial of degree$r$. If$r\ge1$, it does have a non-trivial zero. Then$\Pi':=\Pi(sA+tB)\Pi$is an other product, with rank$\le r-1$. If in addition$\Pi$has$2^{n-r}-1$factors, then$\Pi'$has$2^{n+1-r}-1$factors. After$n$steps, one obtain that the obtains a product has of$2^n-1$factors whose rank is$0$. 1 To my taste, it seems more natural to let$A$and$B$play symmetric role, by asking whether there exists non-trivial factors$s_jA+t_jB$such that $$A\left(\prod_{j=1}^p(s_jA+t_jB)\right)B=B\left(\prod_{j=1}^p(s_jA+t_jB)\right)A.$$ If you pose the question in an algebraically closed field$k$(say,$k=\mathbb C$), then the answer is yes for the following reason: There exist$n$non-zero factors$s_jA+t_jB$such that$\prod_{j=1}^n(s_jA+t_jB)=0$. The proof is by induction over the rank of products$\prod_{j=1}^p(s_jA+t_jB)=0$. Suppose that exists such a product$\Pi$, with rank$r\ge1$. Let us write $$\Pi=\sum_{j=1}^rx_ja_j^T.$$ Then $$\Pi M\Pi=\sum_{i,j=1}^r(a_i^TMx_j)x_ia_j^T.$$ The rank of$\Pi M\Pi$will be less than or equal to$r-1$if$\det(a_i^TMx_j)_{1\le i,j\le r}=0$. When$M=sA+tB$, this writes$H(s,t)=0$where$H$is a homogeneous polynomial of degree$r$. If$r\ge1$, it does have a non-trivial zero. Then$\Pi':=\Pi(sA+tB)\Pi$is an other product, with rank$\le r-1$. After$n$steps, one obtain that the product has rank$0\$.