About finite group generated by 2 elements Suppose (additive) group $G$ is generated by $a$ and $b$ of finite orders
Is it always possible to express every element of $G$ as $ia+jb$ like dihedral group?
 A: The addition 'additive' highly suggests that the group is abelian. On the other hand, the dihedral group you use as an example later on is not abelian. In fact the answer depends on whether or not the group is abelian.
If the group is abelian the answer is yes, almost trivially: if I have an expression a + a + b + a + b + ... I can just change the order to get all the $a$s on the left and I am done.
If the group is not abelian it depends on whether or not I have a way of changing the order. In the dihedral group (which I will write multiplicatively) we have that $b * a = a * b^n$ for some $n$, hence we can use this equality to 'tidy up' any expression of the form $a * b * b * a * a * b * a * b * b...$ by flipping $a$s and $b$s just until all $a$s are on the left. (At which point it is in the form you describe, only I would write it $a^i * b^j$ rather than $ia + jb$.)
However, suppose I have a free product of two cyclic groups, e.g. the group generated by $a$ and $b$ with no other relations than $a^2 = 1$ and $b^3 = 1$, then there is no way to flip the order of $a$ and $b$ and the element $b*a$ for instance can not be expressed in the form $a^n * b^m$.
Of course this group is infinite. You could now wonder if your claim is true for finite non-abelian groups generated by two elements. I expect that the answer is no and that the smallest counterexample is $S_4$, but I would have to think more about it.
