Is it generally true that $\langle a,b \rangle \cong \langle c,d \rangle\Rightarrow \text{ either }|a|=|c|,|b|=|d| \text{ or } |a|=|d|,|b|=|c|$? Background: We are given two groups $G,H$ generated by two elements, say $G=\langle a,b\rangle$ and $H=\langle c,d\rangle$. Further suppose that the orders of $a,b,c,d$ are finite and $\{|a|,|b|\}\neq\{|c|,|d|\}$. Can we conclude that $G\not\cong H$?
Motivation: I have two groups $G$ and $H$, both being non-Abelian and of order 8 (in fact, $G$ is the quaternion group and $H$ is the dihedral group of degree 4), and I know a set of generators for both of the two groups, say $\{a,b\}$ and $\{c,d\}$ correspondingly. I want to show that $G$ is not isomorphic to $H$ since $|a|=|b|=|d|=4$ but $|c|=2$, where $|a|$ is the order of $a\in G$.

No, take $G=H=\mathbb{Z}_4$ which has two sets of generators $\langle 1,2\rangle$ and $\langle 1, 3\rangle$. They satisfy the assumption but the order equalities don't follow.
For non-cyclic case note that if $\langle a, b\rangle$ generates a group then so does $\langle a, ab\rangle$. Now for any integers $n,m,r>1$ there is a (finite) group $G$ and elements $a,b\in G$ such that $|a|=n$, $|b|=m$ and $|ab|=r$. For details see Theorem 1.64 here. That gives you more counterexamples.
