Given two quasisymmetric functions (see definition below) that are not symmetric, must their product also not be symmetric?
From Stanley's Enumerative Combinatorics 2, recall that a function $f\in \mathbb{Q}[[x_{1}, x_{2}, \dots]]$ of bounded degree is said to be quasisymmetric if the coefficient of $x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\dots x_{i_{\ell}}^{\alpha_{\ell}}$ in $f$ is the same as the coefficient of $x_{j_{1}}^{\alpha_{1}}x_{j_{2}}^{\alpha_{2}}\dots x_{j_{\ell}}^{\alpha_{\ell}}$ in $f$ for all $1 \leq i_{1} < \dots <i_{\ell}$ and $1 \leq j_{1} < \dots <j_{\ell}$ and for all positive integer sequences $(\alpha_1,\dots,\alpha_\ell)$ of exponents.
For example $x_{1}x_{2}^{2}+x_{1}x_{3}^{2}+x_{2}x_{3}^{2}$ is a quasisymmetric polynomial in the variables $x_{1}, x_{2}, x_{3}$ that is not symmetric as it does not contain the terms $x_{1}^{2}x_{2}$, $x_{1}^{2}x_{3}$, and $x_{2}^{2}x_{3}$.
I know that the space of quasisymmetric functions and symmetric functions are each closed under multiplication, but I am curious if the space of quasisymmetric functions that are not symmetric is also closed under multiplication.
I have tried several small examples, but I cannot seem to find a counterexample or a proof. I have also experimented with expanding these functions into different bases such as the monomial quasisymmetric basis and Gessel's fundamental quasisymmetric basis and looked at multiplication this way but to no avail.
For a counterexample, let $\,f\,$ be a bivariate quasisymmetric polynomial that is not fully symmetric, and let $\,g(x_1, x_2)=f(x_2, x_1)\,$. Then $\,g\,$ is also quasisymmetric but not fully symmetric, yet their product $\,f(x_1,x_2) \cdot g(x_1, x_2)=f(x_1,x_2) \cdot f(x_2, x_1)\,$ is fully symmetric.