There is a very useful identity, which is,
e^{(a+b)} = e^{a} * e^{b}
However this only applies when 'a' and 'b' commute, so it applies when a or b is a scalar for instance.
The more general case where 'a' and 'b' don't necessarily commute is given by:
e^{c} = e^{a} * e^{b}
where:
c = a + b + a×b + 1/3(a×(a×b)+b×(b×a)) + ...
where:
 × = vector cross product
This is a series known as the BakerCampbellHausdorff formula.
This shows that if when a and b become close to becoming parallel then a×b approaches zero and c approaches a + b so the rotation algebra approaches vector algebra.
Derivation
We will use the formula for an infinite series of an exponent:
e^{(q)} = 


So multiplying two of these together we get:
e^{(a)} *e^{(b)} = 




Now we want to move the second summation outside the first part, but since the first part has already been summed 'n' times we need to substitute n with nm to correct for that:
e^{(a)} *e^{(b)} = 




I don't think what we have done so far has implied any commutation because the operands are still in the same order.