The statement is not true. For a counterexample, let $X$ be a set with at least two elements, let $S={\mathcal P}(X)$ be the power set of $X$, let addition in $S$ equal the union operation, and let multiplication equal intersection. The order relation on $S$ is just the inclusion order.

Let the involution be the function that swaps subsets $\emptyset$ and $X$, but fixes all other elements. Now choose some nonempty proper subset $U$ of $X$ and define $a, b, c\in S$ as follows: $a=U, b=X-U, c=\emptyset$. Then it is easily computed that $ab = \emptyset\leq \emptyset =c$, but $c^*a = a = U\not\leq X-U = b^*$.

The problem here is that the involution is not related to any of the other operations, so one cannot hope to prove $ab\leq c \Leftrightarrow c^*a\leq b^*$. But one can check that this bi-implication is true for the powerset semiring $\langle {\mathcal P}(X); \cup, \cap\rangle$ if you take the involution to be complementation: $U^* = X-U$. If you assume some complementation-like properties for the involution, you will be able to arrange that $ab\leq c \Leftrightarrow c^*a\leq b^*$ holds.