It's not that PIE roots always contain the vowel e, it's that PIE roots don't contain vowels. This is a common misconception, unfortunately aided by the traditions of IE lexicography.
Take a root like lei̯kw- 'leave'. This root is found in:
- e-grade, e.g. Gk. pres. leip-ō
- o-grade, e.g. Gk. pf. le-loip-a
- zero-grade, e.g. Gk. aor. e-lip-on
What this shows is that the vowel (or lack thereof) depends on the grammatical category, not the root. The e in leip-ō is due to the fact that this class of present stems are formed with e, not to anything about the root's lexical entry. The tradition in IE studies is to cite all roots in the e-grade, but it could just as well have been the o-grade (in which case you might now be asking why all PIE roots contain the vowel o). It would be less misleading to cite roots as e.g. l-i̯kw, with no vowels at all, but for historical reasons this isn't how it's done.
The explanation for the existence of some roots with a, at least those which can't plausibly be ascribed to the combination of e with h2, is probably that in late PIE this system was beginning to break down, as it does in all the daughter languages, with consonantal roots starting to give way to vowel-containing stems. But there are very few of these a-roots, i.e. on the whole the consonantal system is still intact.
(The system I'm describing looks rather like the Semitic root system, but there's an important difference, namely that PIE roots contain consonant clusters which can't be broken up: for example you never see a form like li̯ekw-.)
Z
} to represent all the allomorphs of noun plural in English: glasses, bottles, hats have, respectively, the allomorphs /-əz, -z, -s/, all symbolized asZ
, with variants handled automatically by the distribution rules. In both cases, knowing the distribution rules is part of using the root.