Are two octonion algebras with different multiplications isomorphic? Some authors, e.g. Baez, Ward, defined multiplication of octonions by formula
$
(a,b) \cdot^B (c,d)=(ac-db^*, cb+a^*d) \textrm{  for } a,b,c,d\in \mathbb H,
$
some others, e.g. Springer & Veldkamp, N. Jacobson, by
$
(a,b)\cdot^S(c,d)=(ac-d^*b, da+bc^*) \textrm{  for } a,b,c,d\in \mathbb H
$
($a^*$ denotes the conjugate quaternion to $a$).
Which of these multiplications is better?  
Are algebras $\mathbb O=\mathbb H \times \mathbb H$  with multiplications $\cdot^B$ and $\cdot^S$ (and the standard addition and multiplications by real numbers) isomorphic?  
 A: Yes: the isomorphism is given by $(a,b) \mapsto (a,b^*)$.
A: The "S" formula you give corresponds to writing octonions in the form $q+r\ell$ where $q,r\in\mathbb{H}$ and $\ell$ is a unit octonion orthogonal to $\mathbb{H}$: we have
$$(q+r\ell)(q'+r'\ell) = (qq'-r'^*r)+(r'q+rq'^*)\ell$$
which is your "S" formula, written with a more readable notation.  The "B" formula you give corresponds to writing octonions in the form $q+\ell r$: we have
$$(q+\ell r)(q'+\ell r') = (qq'-r'r^*)+\ell(q^*r'+q'r)$$
They can be deduced from one another by using the fact that $q + \ell r = q + r^*\ell$ (a consequence of either formula) which gives the isomorphism you ask for: it's an easy exercise to check that either formula gives the other by applying this isomorphism (e.g., $q(\ell r') = q(r'^*\ell) = (r'^* q)\ell = \ell(q^* r')$ applying the "S" formula for the middle equality).
As to which is best, it really depends whether you prefer to write $\mathbb{O}$ as $\mathbb{H} \oplus \mathbb{H}\ell$ or $\mathbb{H} \oplus \ell\mathbb{H}$.  Personally prefer the former (so, "S"), but that's maybe just because I'm right-handed.
Incidentally, as for how to remember the formula, I think it's better to keep it split as (1) $q(r'\ell)=(r'q)\ell$, (2) $(r\ell)q' = (rq'^*)\ell$ and (3) $(r\ell)(r'\ell) = -r'^*r$ (obviously the purely quaternionic part doesn't need any remembering): the main difficulty is to remember the order of the factors.  It's possible to recover them by remembering that $\ell$ anti-commutes with purely imaginary quaternions and anti-associates with a pair of orthogonal imaginary quaternions, but I think a better mnemonic is that for a unit quaternion $w$, the map fixing $\mathbb{H}$ and taking $r\ell$ to $(wr)\ell$ should be an automorphism of the octonions: this tells us, for example, that $r'$ should be on the left in the formula (1), because we need $q((wr')\ell)$ to be $(wr'q)\ell$, and it similarly constrains (2) and (3) so as to make mistakes impossible.
