Does SHA-1 hash have quasi commutative property? Although I have tested with a program and found negative answer, I want to be sure.
No, $\operatorname{SHA-1}$ does not have any property remotely similar to the quasi-commutative property, that
$$\forall x,\forall  y_1,\forall y_2, f(f(x,y_1),y_2)=f(f(x,y_2),y_1)$$

That's for that natural mapping of $f(x,y)$ to $f(x\|y)$, where $\|$ stands for concatenation. No function (except trivially narrow) can exhibit such property and also aim at being similar to a random function. And the only striking property distinguishing $\operatorname{SHA-1}$ from a random function (beside being $\operatorname{SHA-1}$) is that it has the length-extension property (which allows, given $\operatorname{SHA-1}(x)$ and the length of $x$, to exhibit non-empty $y$ such that $\operatorname{SHA-1}(x||y)$ is known).