If $A$ is compact, $\lambda\ne0$ and $y$ is fixed, is there a unique $x$ with $(\lambda-A)x=y$? If $B$ is a compact linear operator on a Banach space $X$, we know that $\dim\mathcal N(1-B)\in\mathbb N_0$ and $\mathcal R(1-B)$ is closed.
Thus, if $\lambda\ne0$ and $$B:=\frac A\lambda$$ such that $\dim\mathcal N(1-B)=0$, then for all $y\in X$, there is at most one $x\in X$ with $(\lambda-A)x=y$. But can we guarantee that there always is such a $x$?
A bounded operator $T$ is said to be Fredholm if $\text{dim}(\mathcal N(T))&lt;\infty $,  and
$\text{codim}(\mathcal R(T))&lt;\infty $.  In this case the index of $T$ is defined by
$$
  \text{ind}(T)=
  \text{dim}(\mathcal N(T))-\text{codim}(\mathcal R(T)).
  $$
It is a well known result,  sometimes called the Fredholm alternative,  that $I-B$ is a Fredholm
operator with
index zero  for every compact operator $B$.
Thus,  if $\mathcal N(I-B) = \{0\}$,  it follows that $\mathcal R(I-B) = X$ so, yes, the desired $x$ always exists.
Let $ \lambda \ne 0$. If $ \lambda $ is an eigenvalue of $A$, then
$$0&lt; \dim \mathcal{N}(\lambda -A) &lt; \infty,$$
since $A$ is compact. If $y \in \mathcal{R}(\lambda -A)$, then there is $x_0 \in X$ with $(\lambda -A)x_0=y$.
Then we have $(\lambda -A)(x+x_0)  =y$ for all  $x \in  \mathcal{N}(\lambda -A).$