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 $0$.
