Domain theory: Definition of isomorphisms
I think I have found an answer to my question, by proofing that the monotonicity of $f$ and $f^{-1}$ imply the continuity of $f$ and $f^{-1}$.
Please correct me if I made a mistake – I will give you some days to check my proof before I accept my own answer.



Lemma: If a function $f \colon M \to N$ between two cpos is monotonic, the inequality

$$\bigsqcup\{f(m_0), f(m_1), f(m_2), \dots\} \leq f\left(\bigsqcup\{ m_0, m_1, m_2, \dots\}\right)$$

holds for all chains $m_0, m_1, m_2, \dots$. This is well known and easy to show, so I will skip the proof of this.



Let $f \colon M \to N$ be now a bijective function between two cpos and let it, and its inverse $f^{-1}$, be monotonic.
By the above lemma we have already one side of inequality to proof the continuity of $f$. It remains to show that

$$f\left(\bigsqcup\{ m_0, m_1, m_2, \dots\}\right) \leq \bigsqcup\{f(m_0), f(m_1), f(m_2), \dots\}$$

holds for all chains $m_0, m_1, m_2, \dots$ in $M$.
Since $$\bigsqcup\{f(m_0), f(m_1), f(m_2), \dots\} = f\left(f^{-1}\left(\bigsqcup\{f(m_0), f(m_1), f(m_2), \dots\}\right)\right),$$

and both $f$ and $f^{-1}$ are monotonic, the above inequality is equivalent to

$$\bigsqcup\{ m_0, m_1, m_2, \dots\} \leq f^{-1}\left(\bigsqcup\{f(m_0), f(m_1), f(m_2), \dots\}\right).$$

However, again by above lemma we have

$$ f^{-1}\left(\bigsqcup\{f(m_0), f(m_1), f(m_2), \dots\}\right)
\geq \bigsqcup\{f^{-1}(f(m_0)), f^{-1}(f(m_1)), f^{-1}(f(m_2)), \dots\}
= \bigsqcup\{ m_0, m_1, m_2, \dots\}.$$

So $f$ is continuous. You show the continuity of $f^{-1}$ in the same way.