Let me assume that the characteristic of the ground field is different from two. Let me start by replacing your identity by something where the existing symmetries are a bit more apparent. I claim that your identity for the associator (which I will denote simply $(a,b,c)$, dropping $A_*$) is equivalent to the following two identities:
$$
\begin{gather}
(a,b,c)+(b,c,a)+(c,a,b)=0,\\
(a,b,c)+(c,b,a)=0.
\end{gather}
$$
First, they clearly imply your identity: $$0=(a,b,c)+(b,c,a)+(c,a,b)=(a,b,c)+(b,c,a)-(b,a,c).$$ Second, your identity, if we set $a=b=c$, becomes third power associativity $(a,a,a)=0$, and multilinearizing that, we get
$$
(a,b,c)+(a,c,b)+(b,c,a)+(b,a,c)+(c,a,b)+(c,b,a)=0.
$$
Using your identity $(a,b,c)+(b,c,a)=(b,a,c)$ and the identity $(a,c,b)+(c,b,a)=(c,a,b)$ obtained by acting by the transposition $b\leftrightarrow c$, we obtain the identity $(b,a,c)+(c,a,b)=0$, and we use the same calculation as before:
$$0=(a,b,c)+(b,c,a)-(b,a,c)=(a,b,c)+(b,c,a)+(c,a,b).$$
These two identities, in turn, are manifestly equivalent to the system of identities
$$
\begin{gather}
(a,b,c)+(a,c,b)+(b,c,a)+(b,a,c)+(c,a,b)+(c,b,a)=0,\\
(a,b,c)+(c,b,a)=0.
\end{gather}
$$
The first of them defines Lie admissible algebras, and the second defines flexible algebras. Search for papers that include "flexible Lie-admissible algebras" in the title brings 15 matches on MathSciNet, including, for instance, Benkart, Georgia M.; Osborn, J. Marshall. Flexible Lie-admissible algebras.
J. Algebra 71 (1981), no. 1, 11–31. I understand that this class of algebras was studied because it contains the so called "Okubo algebra" arising in one of the constructions of octonions.