Hint $\ $ By the subring test it suffices to verify $\rm\:0-0,\, 0\cdot 0\,\in\, S,\:$ i.e. $\rm \,S\,$ is closed under subtraction and multiplication.
Using your definition of "ring of sets in measure theory", that it is a (nonempty) collection of sets $R$ such that (1) it is closed under union ($\forall A,B\in R$ we have $A\cup B\in R$) and (2) it it is closed under set-theoretic difference ($\forall A,B\in R$ we have $A-B\in R$), we can in fact show $R$ is a commutative ring in the algebra sense, however with respect to the operations symmetric difference and intersection. So first we show that in fact, any ring of sets in measure theory is closed over symmetric difference and intersection:
Lemma. Let $R$ be a ring of set, then for all $A,B\in R$, we have $A\Delta B\in R$ and $A\cap B\in R$. (Here $A\Delta B:=(A-B)\cup(B-A)$ is the operation symmetric difference.)
Pf. The closure of symmetric difference is clear from its definition. And note well that $A\cap B= (A\cup B)-(A\Delta B)$. $\Box$
Now we make the observation: If $R$ is a ring of sets, then $\varnothing\in R$, since for any $A\in R$, we have $A\Delta A=\varnothing \in R$. Further note that for any $A\in R$, we have $\varnothing\Delta A=A$. Hence if we identify the operation $\Delta$ as "addition" and $\varnothing$ as the "additive identity", and every $A\in R$ is its own "additive inverse", we have
Claim. If $R$ is a ring of sets, then $(R,\Delta)$ is an abelian gorup. $\Box$ (Associativity given set-theoretically.)
Now, identify the operation $\cap$ as "multiplication", then with the following
Fact. Intersection distributes over symmetric difference. $\Box$
we have finally:
Proposition. If $R$ is a ring of set (measure theory sense), then it is also closed under symmetric difference $\Delta$ and intersection $\cap$. And that $(R,\Delta,\cap)$ forms a commutative ring (algebra sense). $\Box$
Best Answer
You might apply the fact that if $\phi$ is a ring homomorphism (so respecting the addition and multiplication) from a ring $R$ to a set, then its image $\phi(R)$ is again a ring.
Now let $\phi$ be the map from the normal ring of reals $\mathbb{R}$ to $\mathbb{R}_{>0}$ with your addition $\oplus$ and multiplication $\otimes$, defined by $\phi(a)=e^{a}$, where $e$ is the base of the natural logarithm. Obviously this is a well-defined bijection. One can easily check that $\phi$ is a ring homomorphism. Observe that 0 is mapped to 1 (the neutral element w.r.t. $\oplus$), and the unit 1 is mapped to $e$ (the neutral element w.r.t. $\otimes$). Hence, the two rings are even isomorphic!