About algebras such that $ j_a j_b \in \{ -1,0, +1,j_c,-j_c \}$ for all $a,b$

Question

Consider a commutative unital algebra $A$ of finite dimension $n>3$ over the reals. The product is defined such that elements are generated with real number coefficients $(a_0, \dotsc, a_n) $ for some basis $\{1, i_1, \dotsc, i_n \}$.

Notice $A$ might be non-associative.

Define "F-nilpotent" as there is a nonzero element $x$ in $A$ such that $(x^2) (x^3) = 0$

Conjecture $B$ :

If the algebra $A$ is not "F-nilpotent", then we can always pick a basis $\{1, j_1, \dotsc, j_n \}$ for our algebra $A$, such that $ j_a j_b \in \{ -1,0, +1,j_c,-j_c \}$ for all $a,b$ and some $c$ as a function of $a$ and $b$ ($c = f(a,b) $).

1. How to prove or disprove it ? Any counterexamples ?

2. How to compute such a basis ?

3. How does this relate to power-associativity ?

Looks a bit like group theory or linear algebra so maybe those fields help.

Notice this is trivially true for algebras that are isomorphic to copies of the complex numbers such as for instance the bicomplex numbers.