Why isn't reflexivity redundant in the definition of equivalence relation?

Actually, without the reflexivity condition, the empty relation would count as an equivalence relation, which is non-ideal.

Your argument used the hypothesis that for each $a$, there exists $b$ such that $aRb$ holds. If this is true, then symmetry and transitivity imply reflexivity, but this is not true in general.

No.

The missing condition is sometimes called 'seriality' -- for any x there must be an y such that x R y.

If you add seriality to the symmetry and transitivity you get a reflexive relation again.

Consider the set $S =\{a,b,c\}$, with $a, b$, and $c$ distinct, and the relation $$R = \{(a,b),(b,a),(a,a),(b,b)\} \subset S \times S$$

It's symmetric and transitive but it's not reflexive.

Added 10/28/2018

The argument $aRb \implies bRa \implies aRa$ is compelling except for one thing. What if there is no $b$ such that $aRb$?