This is never true if the kernel of $B$ is symmetric and (say) positive-definite, as follows from the spectral theorem. In this case, $A$ has dense range for $\overline{\operatorname{ran}A} = (\operatorname{ker}A)^\perp=L^2(\mathbb R^3)$, but it is not surjective and $A^{-1}$ is unbounded. That $A$ is injective excludes $0$ from being an eigenvalue of $A$, but $0$ is still in the spectrum of $A$.

Indeed, $\sigma(B)\subseteq [0,1]$ with $1$ belonging to the spectrum of $B$ (since $\|B\|_{L^2\to L^2} =1$) implies that $\sigma(A)\subseteq [0,1]$ with $0$ belonging to the spectrum of $A=I-B$.

**Example.** Take the *Gauss-Weierstrass kernel* $(4\pi)^{-3/2}e^{-|x-y|^2/4}$ as kernel of $B$. There are two possibilities to see that $1$ is not an eigenvalue:

1) It holds $Bf= k\ast f$ with $k(x) = (4\pi)^{-3/2}e^{-|x|^2/4}$. If
$f$ was a (normalized) eigenfunction, then $\bigl(\hat k(\xi)-1\bigr)\,\hat{f}(\xi)=0$ a.e., with $\hat{f}$ being the Fourier transform of $f$. But $\hat k(\xi) = e^{-|\xi|^2}$, so $\hat f(\xi) = 0$ a.e., a contradiction.

2) It holds $B= e^\Delta$, i.e., $B$ is the solution operator of the initial-value problem for the heat equation at time $1$. If $f$ was an eigenfunction of $B$, then $f$ would be an eigenfunction of the generator $\Delta$ (with eigenvalue $0$ as $e^0=1$), which again is a contradiction.