Let $f\colon A\to B$ a morphism of spectra and suppose that both spectra $A$ and $B$ have only one non-zero stable homotopy group $\pi_n$, more precisely $$ \pi_k(A)=0=\pi_k(B) $$ for $k\neq n$. Suppose further that $f$ induces the zero-morphism on $\pi_n$, i.e. $$ \pi_n(f)\colon \pi_n(A)\to \pi_n(B) $$ is the zero-morphism of abelian groups.
Is $f$ necessarily the zero-morphism in the stable homotopy category?
I think this may be true because the heart of the usual t-structure on the stable homotopy category is equivalent to the category of abelian groups, but I am not sure about this.
Surprisingly there are non-zero morphisms of spectra (so called non-zero ghost maps) that induce the zero-map on all stable homotopy groups.
Yes.
If $A$ and $B$ have only one non-zero stable homotopy group they both are Eilenberg-Maclane spectra (shifted by $n$, if you will).
Maps between E-M spectra correspond to (stable) cohomological operations and $[EM(A),EM(A)]_0=\operatorname{Hom}(A,A)$.