The second condition also implies that $A$ is commutative. If $A$ is not commutative then it has an irreducible representation on some Hilbert space $H$ of dimension at least $2$. Find unit vectors $v,w \in H$ with $\langle v, w\rangle = 0$. By Kadison transitivity there exists $x \in A$ with $xv = 0$ and $xw = v$. Then $\langle x^*xv, v\rangle = \|xv\|^2 = 0$ but $\langle xx^*v, v\rangle = \|x^*v\|^2 \neq 0$ because $\langle x^*v, w\rangle = \langle v, xw\rangle = 1$. So $xx^* \leq kx^*x$ is impossible.
(The second part of the question doesn't really make sense because if $k \in A$ then $kx^*x$ will not be positive in general. You could ask about $xx^* \leq x^*kx$, but this would imply $xx^* \leq \|k\|x^*x$ and therefore $A$ must be commutative by the scalar case.)