I'm nor sure whether the following works:

Take an operator on a Banach space whose image is dense, whose spectrum is $\{0\}$ but that has no kernel, for example $$ T(f)(x)=\int_x^1f(y)dy $$ acting on $L^2([0,1])$.

Then its restriction to the dense subspace $\bigcap_n Im(T^n)$ should have the property you desire.

Take an operator on a Banach space whose image is dense, whose spectrum is $\{0\}$ but that has no kernel, for example $$ T(f)(x)=\int_x^1f(y)dy $$ acting on $H:=L^2([0,1])$.

Then its restriction to the dense subspace $D:=\bigcap_n Im(T^n)$ should have the property you desire.
First of all, $D$ is dense in $H$ -- see the comments below by Alexander Shamov.

Clearly, $T|_D$ is invertible.

So all that remains to be checks is that for $\lambda\not=0$, the operator $(T-\lambda)$ is invertible on $D$. Recall that $(T-\lambda)^{-1}$ makes sense on $H$. The trick is to note that $(T-\lambda)^{-1}$ preserves $D$. Indeed it preserves each subspace $Im(T^n)$: If $x\in Im(T^n)$ write it as $x=T^ny$ and then we have $(T-\lambda)^{-1}x=(T-\lambda)^{-1}T^ny=T^n(T-\lambda)^{-1}y\in Im(T^n)$. QED

I'm nor sure whether the following works:

Take an operator on a Banach space whose spectrum is $\{0\}$ but that has no eigenvectors, for example $$ T(f)(x)=\int_x^1f(y)dy $$ acting on $L^2([0,1])$.

Then its restriction to the dense subspace $\bigcap_n Im(T^n)$ should have the property you desire.

I'm nor sure whether the following works:

Take an operator on a Banach space whose image is dense, whose spectrum is $\{0\}$ but that has no kernel, for example $$ T(f)(x)=\int_x^1f(y)dy $$ acting on $L^2([0,1])$.

Then its restriction to the dense subspace $\bigcap_n Im(T^n)$ should have the property you desire.