A magma is alternative if for all elements $x$ and $y$, we have $(xx)y = x(xy)$ and $y(xx) = (yx)x$. A magma is flexible if for all elements $x$ and $y$ we have $x(yx) = (xy)x$. Both of these are weaker forms of associativity.
This wiki page currently states that both alternative algebras and alternative magmas are flexible.
For alternative algebras, this follows from Artin's theorem, which states that any two elements of an alternative algebra generate an associative subalgebra.
But, why must alternative magma's be flexible?
This is false. For instance, let $A$ be the free magma on two elements $x$ and $y$ (consisting of parenthesized words in $x$ and $y$), and let $\sim$ be the equivalence relation defined by $(xx)x\sim x(xx),$ $(yy)y\sim y(yy),$ $(xx)y\sim x(xy),$ $(yy)x\sim y(yx),$ $y(xx)\sim(yx)x,$ $x(yy)\sim(xy)y,$ and all words of length $4$ or more are equivalent. It is easy to verify that this is a congruence relation (there is basically nothing to check since all nontrivial instances would involve words of length $4$ or more), and so $A/{\sim}$ is a magma. It is alternative: every instance of alternativity involving just $x$ and $y$ was enforced by $\sim$, and every other instance gives words of length $4$ or more. However, it is not flexible since $x(yx)\neq (xy)x$.