If $T$ is invertible and $K$ is compact, is $0\ne\lambda\mapsto(\lambda T+K)^{-1}$ holomorphic? Let $X$ be a Banach space, $T\in\mathfrak L(X)$ be bijective, $K\in\mathfrak L(X)$ be compact and $$B:\mathbb C\to\mathfrak L(X)\;,\;\;\;\lambda\mapsto\lambda T+K.$$

Can we show that $B(\lambda)$ is bijective for all $\lambda\in\mathbb C\setminus\{0\}$ and $$\mathbb C\setminus\{0\}\to\mathfrak L(X)\;,\;\;\;\lambda\mapsto B(\lambda)^{-1}$$ is holomorphic?

This is clearly, if correct, an application of the analytic Fredholm theorem.
The trick is clearly to write $$B(\lambda)=\lambda T\left(1+\frac1\lambda T^{-1}K\right)\tag1\;\;\;\text{for all }\lambda\in\mathbb C\setminus\{0\}.$$ $\mathbb C\ni\lambda\mapsto\lambda T$ and $\mathbb C\setminus\{0\}\ni\lambda\mapsto\frac1\lambda T^{-1}K$ are trivially holomorphic. Moreover, $T^{-1}K$ is compact.
So, the only missing piece is that we need to show that $B(\lambda)$ is bijective for at least one $\lambda\in\mathbb C$. But why is that the case?

You seem to assume invertibilty of $T$. In that case $T+\frac 1 n K \to T$ in operator norm. Since the set of all invertble operators is open it folows that $T+\frac 1 n K$ is invertible  for large enough $n$ and this implies that $nT+K$ is invertible for such $n$.
