Are there axioms for dividing by zero that don't destroy math?

Regarding a foolproof method: Produce a model of the set of sentences (properties) you wish to hold. A theory $T$ is consistent if and only if there exists a model of $T$. (See, for example, the second form of completeness mentioned in the answer to Question about the proof of consistency iff satisfiability of a theory)

I interpret your question as being about a signature $(+, \cdot, 0, 1, \omega)$. Informally, what "produce a model" means in this context is to find a set $X$ and two binary operations $+$,$\cdot$, and a mapping of the constant symbols $0$,$1$,$\omega$ into $X$ that satisfy whatever list of sentences you have in mind. As mentioned in the comments, it's not hard to produce examples if you don't require distributivity.

For example, consider the set $\{a,b\}$ with $+$ exhaustively given by $a + a = a$, $a + b = b$, $b + a = b$, $b + b = a$ and $\cdot$ exhaustively given by $a \cdot a = b$, $a \cdot b = a$, $b\cdot a = a$, and $b \cdot b = b$. Interpreting $0$ as $a$, $1$ as $b$, and $\omega$ as $a$, we see $\omega \cdot 0 = 1$ in this model and I believe we can hand-check all commutative ring axioms except for distributivity. (This is a disguised version of XOR as $+$ and NOT-XOR as $\cdot$.) If I haven't messed up, this implies the commutative ring axioms without distributivity together with a constant $\omega$ satisfying $\omega \cdot 0 = 1$ won't produce an inconsistency.