You still need to show that:
$\oplus_S$ has a neutral element.
$\oplus_S$ is associative.
If $0_S$ is the aforementioned additive neutral element, that $\oplus_S$ has opposites, id est, that for all $a$ there is $b$ such that $a\oplus_S b= 0_S$ and $b\oplus_S a=0_S$.
If your definition of ring requires it (most of them do), that $\otimes_S$ has a neutral element as well.
Some would say that you also need to check "closedness under operations", which would technically mean showing that $$\forall a,b\in S,\ (a\oplus_S b\in S\wedge a\otimes_S b\in S)$$
This is reminescent of one of the verifications that you need to do when you check that something is a subring of something else. In this case, it would be an abuse of terminology, because clearly $\oplus_S$ and $\otimes_S$ should already be functions with codomain $S$. What should be checked beforehand in this case is that $\oplus_S$ and $\otimes_S$ are two well-defined functions $S\times S\to S$.
What does that mean? Describing it in general is a bit abstract, but somewhere you have a predicate which describes what property, say, $a\oplus_S b$ is meant to satisfy with repsect to $a,b\in S$. Call that property $P(a,b;y)$ (and the definition somehow goes like "$a\oplus_S b$ is the one element satisfying $P(a,b; a\oplus_Sb)$" ). Then you must prove that, for any $a,b\in S$ there is exactly one element $y$ in $S$ such that $P(a,b;y)$ is true. Id est: $$\forall a,b\in S,((\exists y\in S,\ P(a,b;y))\wedge (\forall w,y\in S, (P(a,b;w)\wedge P(a,b;y)\longrightarrow w=y)))$$