Is every commutative simple ring a field, even if we don't assume it to be unital a priori?
What about the two-element ring with zero multiplication?
Note that the in the setting of nonunital rings, a ring $R$ is called simple if its multiplication is not trivial (i.e., $R^2\neq0$) and it has no ideals other than $0$ and $R$. Hence we have two possible answers here:
We allow for trivial multiplication. If $R$ has trivial multiplication then the ideals are exactly the subgroups of $(R,+)$, so $R$ is simple iff $(R,+)$ has no nontrivial proper subgroups, hence it is $0$ or a cyclic group of prime order. Therefore $R$ is $0$ or $(\mathbb Z_p,+)$ with trivial multiplication for some prime $p$.
We restrict to the usual definition of simplicity given above. Then (as I show below) $R$ must have identity element, so it is a simple commutative unital ring, hence a field by the usual proof.
Proposition. A simple commutative ring is unital.
Proof: Since $R^2\neq0$ there is $0\neq a\in R$ such that $Ra=R$, as $Ra$ is a nonzero ideal. That is, $R$ is generated by a single element. Let us see that we can pick an idempotent generator. Since $R=Ra$, there is $e\in R$ such that $a=ea$, whence $ea=e^2a$, i.e., $(e-e^2)a=0$. Suppose $e$ is not idempotent; then $e^2-e\neq0$, so $R=R(e^2-e)$ and there is $b\in R$ such that $e=b(e^2-e)$, so $a=ea=b(e^2-e)a=0$, a contradiction. Hence $e^2=e$. Now let us see that $e$ is actually the identity element of $R$. Let $x\in R$ and suppose $ex-x\neq0$. Then again $R=(ex-x)R$, so there is $y$ (depending on $x$) such that $e=(ex-x)y$, so $e=e^2=e(ex-x)y=(ex-ex)y=0$, a contradiction. Therefore $ex=x$ for all $x\in R$.
Remark: A generalization of the previous proposition is: Let $R$ be a (not necessarily unital nor commutative) semiprime ring and $I$ be a minimal left ideal of $R$. Then there is an idempotent $e\in I$ such that $I=Ie=Re$.