If I've understood your question correctly, you are asking whether two simple groups can have unitally order isomorphic representation rings (where the cone for the partial ordering is given by the actual characters, generated additively by the irreducible characters). There is a partial result available, and group theorists can probably tell us if this is sufficient.

If we view the representation ring of the finite group $G$, $R(G)$, as a partially ordered ring with $1$ (with positive cone generated additively by the irreducible characters), then we can recover the cardinality of $G$ from it. [That is, if $R(G) $ and $R(G_1)$ are isomorphic as unital partially ordered rings, then $G$ and $G_1$ have equal cardinality; for trivial reasons (the rank, as an abelian group), they also have the same number of conjugacy classes, and the same number of irreducible characters.]

Either by the Perron-Frobenious theorem or otherwise, there is a unique positive ring homomorphism $t:R(G) \to {\bf R}$ sending the trivial character to $1$, and this is simply evaluation of the virtual character at $1$, the dimension. Again by the Perron theorem, up to scalar multiple, there is a unique common eigenvector (viewing a virtual character as an endomorphism of $R(G)$ by multiplication) for $t$, specifically, a positive real multiple of the regular representation character, call it $\chi$. In particular, $\chi$ is characterized by being the unique character of $G$ that is a common eigenvector, belongs $R(G)^+$, and has minimal degree (valuation at $t$) among those satisfying the first two properties.

It follows that $t(\chi)$ is an invariant of $R(G)$, and of course, $t(\chi) = \chi(1) = |G|$. With a little more work, the set with multiplicities consisting of the degrees of irreducibles is also an invariant of $R(G)$. [Irreducibles have to be mapped to irreducibles under an order isomorphism, and evaluate at $t$.]

So if $R(G)$ and $R(G_1)$ are isomorphic as unital partially ordered rings, then $|G| = |G_1|$, they have the same number of irreducibles, and the degrees of their irreducibles are equal (as multisets); of course, the last also implies cardinalities are equal. For simple groups, I don’t know whether this is enough to distinguish them.

As is well known (and probably what motivated the restriction to simple groups), the two interesting groups of order 8 ($D_4$ and the quaternion group) cannot be distinguished by their representation rings.

If I've understood your question correctly, you are asking whether two simple groups can have unitally order isomorphic representation rings (where the cone for the partial ordering is given by the actual characters, generated additively by the irreducible characters). There is a partial result available, and group theorists can probably tell us if this is sufficient.

If we view the representation ring of the finite group $G$, $R(G)$, as a partially ordered ring with $1$ (with positive cone generated additively by the irreducible characters), then we can recover the cardinality of $G$ from it. [That is, if $R(G) $ and $R(G_1)$ are isomorphic as unital partially ordered rings, then $G$ and $G_1$ have equal cardinality; for trivial reasons (the rank, as an abelian group), they also have the same number of conjugacy classes, and the same number of irreducible characters.]

Either by the Perron-Frobenious theorem or otherwise, there is a unique positive ring homomorphism $t:R(G) \to {\bf R}$ sending the trivial character to $1$, and this is simply evaluation of the virtual character at $1$, the dimension. Again by the Perron theorem, up to scalar multiple, there is a unique common eigenvector (viewing a virtual character as an endomorphism of $R(G)$ by multiplication) for $t$, specifically, a positive real multiple of the regular representation character, call it $\chi$. In particular, $\chi$ is characterized by being the unique character of $G$ that is a common eigenvector, belongs to $R(G)^+$, and has minimal degree (valuation at $t$) among those satisfying the first two properties.

It follows that $t(\chi)$ is an invariant of $R(G)$, and of course, $t(\chi) = \chi(1) = |G|$. With a little more work, the set with multiplicities consisting of the degrees of irreducibles is also an invariant of $R(G)$. [Irreducibles have to be mapped to irreducibles under an order isomorphism, and evaluate at $t$.]

So if $R(G)$ and $R(G_1)$ are isomorphic as unital partially ordered rings, then $|G| = |G_1|$, they have the same number of irreducibles, and the degrees of their irreducibles are equal (as multisets); of course, the last also implies cardinalities are equal. For simple groups, I don’t know whether this is enough to distinguish them.

As is well known (and probably what motivated the restriction to simple groups), the two interesting groups of order 8 ($D_4$ and the quaternion group) cannot be distinguished by their representation rings.