Question regarding the kernel of $\Lambda:\operatorname{SL}(2)\rightarrow \operatorname{Isom}(\Bbb H_2)$

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Let $\Lambda:\operatorname{SL}(2)\rightarrow \operatorname{Isom}(\Bbb H_2)$ be defined by $\Lambda(g)(z)=\frac{g_{(1,1)}z+g_{(1,2)}}{g_{(2,1)}z+g_{(2,2)}},g\in SL(2),z\in \Bbb H_2$.

I need to check if $\ker(\Lambda)=\{\lambda I_2|\lambda \in \Bbb R \}$. Unless I am missing something, isn't $\lambda I_2 \notin \operatorname{SL}(2)$? (Because $\det(\lambda I_2)$ not necessarily equal to $1$)

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You're correct: By definition $\ker \Lambda$ is a subset of the domain, so the assertion is wrong. Note, however, that $\det(\lambda I_2) = \lambda^2$, so $\det(-I_2) = 1$ and indeed $-I_2 \in \ker \Lambda$.