In a lot of algorithms using trees, we need the property that when folding $2^n$ elements with some operator $+$, we can do the first half of $2^{n-1}$ elements and the second half independently. That is $$((x_1+x_2)+x_3)+x_4 = (x_1+x_2)+(x_3+x_4)$$ which then implies by repeated application $$ \begin{align*} ((((((x_1+x_2)+x_3)+x_4)+x_5)+x_6)+x_7)+x_8 &= (((((x_1+x_2)+x_3)+x_4)+x_5)+x_6)+(x_7+x_8) \\&= ((((x_1+x_2)+x_3)+x_4)+(x_5+x_6))+(x_7+x_8) \\&= ((x_1+x_2)+(x_3+x_4))+((x_5+x_6)+(x_7+x_8)) \end{align*} $$, and so on for larger $n$.
Clearly, associativity is sufficient, but I wonder if you can think of any structures with an operator that only satisfies this weaker condition? (Notice that it is only defined for folds with length $2^n$ for some $n$). I can't think of any, but I don't see why not?
Update: I guess for any group the two notions are the same, since we can write $$(x_1+x_2)+x_3 = ((0+x_1)+x_2)+x_3 = (0+x_1)+(x_2+x_3) = x_1+(x_2+x_3)$$, so maybe the answer is negative.