First of all, we need to clarify just what is a given in this question; I normally take an "integral domain" to be a unital commutative ring $R$ with no zero divisors; that is, $R$ is possessed of a multiplicative identity $e$ such that $ea = ae = a$ for $a \in R$; and if $a, b \in R$, then $[ab = 0] \Longrightarrow [a = 0 \vee b = 0]$; as far as I am aware, this definition is currently widely accepted; but the ubiquity of variants in use cannot be denied. For example, apparently a notion of "integral domain" which doesn't require the existence of $e$ but does demand that $[ab = 0] \Longrightarrow [a = 0 \vee b = 0]$ is still widely accepted. Therefore an attractive feature of the following argument, based as it is upon functions $f_r:R \to R$, $f_r(s) = rs$, is that it allows us to establish both the existence of a multiplicative identity and of multiplicative inverses concurrently. So we may as well drop any assmuption that $R$ need have a unit.
It works like this:
If $f_r:R \to R$ is given by
$f_r(s) = rs, \tag 1$
with $r \ne 0$, then $f_r$ is manifestly injective, since
$f_r(s_1) = f_r(s_2) \Longleftrightarrow rs_1 = rs_2 \Longleftrightarrow r(s_1 - s_2) = 0 \Longleftrightarrow s_1 - s_2 = 0 \Longleftrightarrow s_1 = s_2, \tag 2$
where we have used the fact that $R$ is an integral domain to infer that
$r(s_1 - s_2) = 0 \Longleftrightarrow s_1 - s_2 = 0 \tag 3$
in (2); since $R$ is finite and $f_r$ is injective, $f_r$ is also surjective and thus there exists $e \in R$ with
$re = f_r(e) = r; \tag 4$
$e$ is in fact unique in accord with (2). That $f_r$ is surjective further implies that, for any $s \in R$, there is some $t \in R$ with
$tr = rt = f_r(t) = s; \tag 5$
then
$se = (tr)e = t(re) = tr = s, \tag 6$
which shows that the element $e$ is in fact a multiplicative unit for all of $R$; thus $R$ is unital, and we may once again invoke he surjectivity of $f_r$ to affirm the existence of some $r' \in R$ with
$rr' = f_r(r') = e; \tag 7$
thus $r' = r^{-1}$ is a multiplicative inverse of $r$.
We can obviously define a similar function $f_s$ for any non-zero $s \in R$; the identity element $e_s$ so obtained will in fact be $e$ since in the usual manner we may write
$e_s = e_s e = e; \tag 8$
furthermore, $s \ne 0$ implies $f_s$ is surjective, as above, and thus we establish the existence of $s^{-1}$:
$ss^{-1} = s^{-1} = e. \tag 9$
We have now established that $R$ has a multiplicative identity element $e$ and a multiplicative inverse $r'$ for every element $r \in R$; thus $R$ meets all the requirements to attain the status of field.
As far as our OP Faust's closing questions are concerned:
1.) Yes, $r, s \in R$.
2.) Yes, $f_r$ being injective/surjective is, since $R$ is finite, exactly equivalent to $f_r$ being a permutaion of the elements of $R$. That each such $f_r$ is a permutaion of $R$ is the key idea behind this proof.
3.) The map $f_r$ is really a just convenient way of keeping track of the multiplicative action of $r$ on $R$; a notational/conceptual covenience, if you will. The essential thing is that multpilication by any nonzero element of $R$ generates a permutation of this finite set. We can't simply take $f$ to be a homomorphism to a field since, at the beginning of the argument at least, we haven't yet shown a field is involved. Until we've shown otherwise, we don't even know if there is a field with the same number of elements as $R$.
