Consider a real vector space $V^{(1)}\otimes V^{(2)}$ where $\otimes$ is the tensor product. Product vectors are of the form $v_1\otimes v_2$ where $v_1\in V^{(1)}$, $v_2\in V^{(2)}$, anything else is a non-product vector. I conjecture the following
I have a set of product vectors $a_i\otimes b_i$, $i=1,\dots,n$ that sum to make a product vector $A\otimes B$, $A\otimes B=\sum_i \alpha_i a_i \otimes b_i$ where $\alpha_i$ are real, non-zero constants and there are no repetitions of $a_i$. All of these product vectors must have $b_i= \lambda \, B$ $\forall$ $i$, where $\lambda$ is some non-zero real.
I believe this is true, but am not sure how to proceed. For example, if $n=2$, then $A\otimes B=a_1\otimes b_1 + a_2 \otimes b_2$ is not a product vector unless $a_1=a_2$ or $b_1 =b_2$, and then it is clear that either $a_1=a_2=A$ or $b_1=b_2=B$
Since, $X\otimes Y = P\otimes Q + A$ then yor statement:
is valid (by definition). But the rest are not. Take for example $\mathbb{R}^4$, and a basis $\mathbf{e_1,e_2,e_3,e_4}$. Let $$ A=\mathbf{e_1\otimes e_2+e_3\otimes e_4} $$ Then $A$ has none of the forms: $P\otimes (\dots )$ or $(\dots )\otimes Q$ or $X\otimes (\dots )$ or $(\dots )\otimes Y$
Thus, your conjecture is not true in general.