According to the answers to this question, if $R$ is a ring with no zero divisors, we can define a ring of fractions of $R$ if $R$ satisfies the Ore condition and that ring will be a skew-field. What would happen if $R$ is an algebra over a field is? Say $A$ is an division algebra over a field that is not necessarily associative and may have zero divisors. Is there a condition like the Ore condition that guarantees that an algebra of fractions can be defined?
Can we not just adjust the Ore condition to make it suitable for algebras? I'm not sure if associativity and having no zero-divisors affect the fact that we can extend a ring to a ring of fractions.
The (right or left) Ore conditions are already defined for any multiplicative subset in any ring.
Using them you can define a ring of fractions by considering the subset of regular elements, of the conditions are satisfied.
Allowing zero divisors into the set does not make any sense, because if you try to make an inverse for them, the resulting ring will collapse to zero.