How to derive this from the universal property of the ring of fractions?

Question

(All the involved rings are commutative with identity.)

The "universal" definition that I follow is this:

Let $S$ be a multiplicative subset of a ring $A$ which also contains $1_A$. Then a ring of fractions of $A$ w.r.t. $S$ is a ring $R$ together with a ring homomorphism $i\colon A\to R$ such that the following hold:

1. $i(S)\subseteq R^*$.
2. Any ring homomorphism $A\to B$ with the image of $S$ in $B^*$ factors through $R$ uniquely via $i$.

In the following, I will denote $i(a)\, i(s)^{-1}$ by $a/s$ for the sake of brevity (having fixed an $(R, i)$).

Using just the above definition, I can deduce the usual properties:

1. $R$ is unique up to a unique isomorphism.
2. $R = \{a/s : a\in A, s\in S\}$.

Questions:

1. Is the condition $i(A)\subseteq R^*$ actually necessary in the definition?
2. Can I also deduce that $i(a) = 0$ implies $as = 0$ for some $s\in S$? Or should I posit it in the definition? (This is crucial to deduce the usual fact that $a/s = b/t \iff (at - bs)u = 0$ for some $u\in S$.)

Answer 1

Question 1: Yes, this is necessary. Otherwise, we could take $R = A$ and $i = 1_A$.

Question 2: I’m not sure how you would deduce that directly. An indirect route would be defining $S^{-1} A$ in the ordinary way, and taking the induced map $f : R \to S^{-1} A$. Then if $i(a) = 0$, we have $f(i(a)) = \frac{a}{1} = f(0) = 0$, and therefore there exists $s$ such that $as = 0$.

Guessing the right definition of $S^{-1}A$ is not totally trivial. You can certainly establish that if there exists $s$ such that $s (ad - cb) = 0$, then $\frac{a}{b} = \frac{c}{d}$. You would then have to intuit that this is a necessary condition as well.