Is $(Id-\Delta)\colon H^2(\Omega)\to L^2(\Omega)$ bijective?

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Is

$$(Id-\Delta)\colon H^2(\Omega)\to L^2(\Omega),$$

with Neumann or Dirichlet boundary conditions, bijective? I know that this holds, but why? ($\Omega$ is in $\mathbb{R}^n$ with $n=2,3$ and smooth boundary)

My first Try: $(\lambda-\Delta)$ is bijective for $\lambda<0$ since $\rho(\Delta)\subset \mathbb{R}_{\geq 0}$ but is is it possible that 1 is in $\rho(\Delta)$ too?

My second Try: $-\Delta$ is bijective. Does my claim hold for an easier reason than above?

Sry if there is some bad english and thanks for help.

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Note that if $(I- \Delta) u =0$, then

$$0 = \int_\Omega u(I-\Delta)u dx = \|u\|_2^2 + \|\nabla u\|_2^2.$$

Thus $u =0$. So $I-\Delta$ is injective. Since $I-\Delta$ is Fredholm of index zero, this implies that $I-\Delta$ is also surjective.