Let $A$ be a commutative ring with unity and let $S$ be a multiplicative subset of $A$. The universal property states that given any commutative ring $B$ and ring homomorphism $f:A\to B$ such that $f(s)$ is a unit in $B$ for all $s\in S$, there exists a unique ring homomorphism $\varphi:S^{-1}A\to B$ such that $f=\varphi\circ\iota$, where
$S^{-1}A$ denotes the ring of fractions of $A$ over $S$ and
$\iota:A\to S^{-1}A:a\mapsto a/1_A$ is the canonical map.
Furthermore, the map $\varphi$ satisfies that $$\varphi(a/s)=f(a)f(s)^{-1},\quad\forall a\in A,~s\in S.$$
Here I want to study the injectivity and surjectivity of $\varphi$ with respect to $f$. My conclusion is that $\varphi$ is surjective (resp. injective) if and only if $f$ is. Here are my proofs:
Since $f=\varphi\circ\iota$, it is clear that $\varphi$ is surjective if and only if $f$ is.
For every $s\in S$, by assumption, its image $f(s)$ is always a unit, hence nonzero. It then follows that for every $a\in R$ and $s\in S$, $$ \varphi\left(\frac{a}{s}\right)=0_B\iff f(a)f(s)^{-1}=0_B\iff f(a)=0_B\iff a\in\ker(f). $$ Therefore, we indeed have $$\ker(\varphi)=\ker(f)^e=\{a/s\mid a\in\ker(f),~s\in S\},$$ where $\ker(f)^e$ is the extension of $\ker(f)$ under $\iota$. Hence $\ker(\varphi)=\langle 0\rangle$ if and only if $\ker(f)=\langle 0_A\rangle$, namely $\varphi$ is injective if and only if $f$ is.
I am not so confident about what I have discovered so far, so it would be nice if anyone could check it, thanks!
Corrections. The fact is the injectivity and surjectivity of $f$ would imply them of $\varphi$, but not the converse.
If $f$ is surjective, then necessarily $\varphi$ is, but the converse is naturally false generally.
If $\ker(f)=\langle 0_A\rangle$, one can immediately see that $\ker(\varphi)=\langle 0\rangle$. However, if $\ker(\varphi)=\langle 0\rangle$, it means that $a/s=0$ for all $a\in\ker(f)$ and $s\in S$. Note that for every $a/s\in S^{-1}A$, $$\frac{a}{s}=0\iff\exists u\in S:au=0_A,$$ so we can see that every element in $\ker(f)$ is annihilated by some element in $S$, but still could be nonzero. Nevertheless, if every element in $S$ is not a zero divisor, we shall necessarily have $a=0_A$ whence $\ker(f)=\langle 0_A\rangle$ would hold.
As you note in your corrections, the surjectivity of $f$ implies that of $\varphi$, but not conversely. An easy example to see that you can have $\varphi$ surjective but $f$ not surjective is to take a domain $A$ that is not a field, take $S=A-\{0\}$, and consider $f=\iota\colon A\to S^{-1}A$. Then $\varphi=\mathrm{id}_{S^{-1}A}$ is a bijection, but $f$ is not surjective (since $S^{-1}A$ is the field of fractions of $A$, and this is not equal to $A$).
You have that $\varphi$ is surjective if and only if $B$ is generated, as a subring, by $f(A)$ and $\{f(s)^{-1}\mid s\in S\}$. Though this is a bit of a silly condition, as it essentially translates into "$\varphi$ is surjective if and only if the image of a generating set for $S^{-1}A$ is a generating set for $B$."
For a counterexample in the injectivity clause (that if $\varphi$ is one-to-one, then $f$ is one-to-one) consider $A=\mathbb{Z}/6\mathbb{Z}$ (I will represent the residue classes as $0,1,2,3,4,5$), and let $S=\{2,4\}$. Now let $f\colon A\to \mathbb{Z}/3\mathbb{Z}$ be the map "reduction modulo $3$; that is, $0,3\mapsto 0$, $1,4\mapsto 1$, $2,5\mapsto 2$. Note that the images of $2$ and $4$ are units; and that $f$ is not one-to-one.
What is $S^{-1}A$? Well, $\frac{a}{2}=\frac{b}{2}$ if and only if there exists $s'\in S$ such that $2s(a-b)=0$; since $s$ is not divisible by $3$, this can only happen is $a\equiv b\pmod{3}$. Likewise for $\frac{a}{4}=\frac{b}{4}$. And $\frac{a}{2}=\frac{b}{4}$ if and only if there exists $s\in S$ such that $2s(2a-b)=0$, so we need $2a\equiv b\pmod{3}$. Thus, we have that $S^{-1}A$ has three elements: $$\begin{align*} 0_{S^{-1}A} &= \frac{0}{2} = \frac{0}{4} = \frac{3}{2} = \frac{3}{4}\\ 1_{S^{-1}A} &= \frac{2}{2} = \frac{5}{2} = \frac{1}{4} = \frac{4}{4}\\ -1_{S^{-1}A} &= \frac{1}{2} = \frac{4}{2} = \frac{2}{4} = \frac{5}{4}. \end{align*}$$ So the induced map $\varphi$ is just the map sending $0_{S^{-1}A}$ to $0$, $1_{S^{-1}A}$ to $1$, and $-1_{S^{-1}A}$ to $2$. This is one-to-one, even though $f$ was not one-to-one.
When will $\varphi$ be one-to-one?
Proposition. The induced map $\varphi$ is one-to-one if and only if $$\ker(f) = \{a\in A\mid \exists s\in S\text{ such that }as=0_A\}.$$
Proof. Regardless of any condition on $\varphi$, we have $$\{a\in A\mid\exists s\in S\text{ such that }as=0_A\}\subseteq \ker(f).$$ Indeed, let $a$ be in the set in the left hand side. Then $\frac{as}{s}=\frac{0}{s}$, so $$0_B = \varphi\left(\frac{0}{s}\right) = \varphi\left(\frac{as}{s}\right) = f(a)f(s)f(s)^{-1} = f(a),$$ so $a\in\ker(f)$.
Suppose first that $\ker(f)$ equals the set, and we show that $\varphi$ is one-to-one. If $\varphi(\frac{a}{t})=f(a)f(t)^{-1}=0_B$, then multiplying by $f(t)$ we get that $f(a)=0$. Therefore, there exists $s\in S$ such that $as=0$. Then $\frac{a}{t} = \frac{as}{ts} = \frac{0}{ts} = 0_{S^{-1}A}$. Thus, $\varphi$ is one-to-one.
Conversely, if $\varphi$ is one-to-one, and $f(a) = 0$, then given $t\in S$ we have $\varphi(\frac{a}{t})=0$. Therefore, $\frac{a}{t}=\frac{0}{t}$, so there exists $s'\in S$ such that $s'(at-0t) = (s't)a = 0$. Therefore, $s=s't\in S$ is such that $sa=0$. Thus, $\ker(t)$ is contained in the given set. As we always have the other inclusion, we are done. $\Box$