You have to be careful about what you mean by "positive (semi-)definite" in the case of non-Hermitian matrices. In this case I think what you mean is that all eigenvalues are
positive (or nonnegative). Your statement isn't true if "$A$ is positive definite" means $x^T A x > 0$ for all nonzero real vectors $x$ (or equivalently $A + A^T$ is positive definite). For example, consider
$$ A = \pmatrix{ 1 & 2\cr 2 & 5\cr},\ B = \pmatrix{1 & -1\cr -1 & 2\cr},\
AB = \pmatrix{-1 & 3\cr -3 & 8\cr},\ (1\ 0) A B \pmatrix{1\cr 0\cr} = -1$$
Let $A$ and $B$ be positive semidefinite real symmetric matrices. Then $A$ has a positive semidefinite square root, which I'll write as $A^{1/2}$. Now $A^{1/2} B A^{1/2}$ is symmetric and positive semidefinite, and $AB = A^{1/2} (A^{1/2} B)$ and $A^{1/2} B A^{1/2}$ have the same nonzero eigenvalues.
First, one can argue that non-symmetric positive definite matrices are pathological, in the sense that when you move to the complex case all positive definite matrices are hermitian.
For a non-symmetric positive definite matrix you can say little more than the fact that it has positive eigenvalues. You don't have many of the nice properties that symmetry adds. For instance, without symmetry you don't even have that the singular values agree with the eigenvalues, nor diagonalizability.
Edit: here is why in the complex case, positive semidefinite implies hermitian. Actually, the proof implies that in the complex case $A$ is hermitian if and only if $x^*Ax\in\mathbb R$ for all $x$.
Assume $x^*Ax\in\mathbb R$ for all $x$. then
$$
\mathbb R\ni(y+\alpha x)^*A(y+\alpha x)=y^*Ay+\overline\alpha\,x^*Ay+\alpha\,y^*Ax+|\alpha|^2\,x^*Ax.
$$
As this expression is real, it equals its complex conjugate
$$
y^*Ay+\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y+|\alpha|^2\,x^*Ax.
$$
So
$$
\overline\alpha\,x^*Ay+\alpha\,y^*Ax=\alpha\,y^*A^*x+\overline\alpha\,x^*A^*y.
$$
Taking first $\alpha=1$ and then $\alpha=i$, we get
$$
x^*Ay+y^*Ax=y^*A^*x+x^*A^*y,
$$
$$
-i\,x^*Ay+i\,y^*Ax=i\,y^*A^*x-i\,x^*A^*y.
$$
Multiplying the first equation by $i$ and adding, we get
$$
2i\,y^*Ax=2i\,y^*A^*x.
$$
As this works for any $x,y$, we deduce that $A=A^*$.
Best Answer
For any square matrices $X$, $Y$, the matrices $XY$ and $YX$ have the same eigenvalues ( in fact, the same characteristic polynomial). So you can apply you method even if $A$ is not invertible.