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Fix a closed $T$ and an arbitrary closed target $M$. If $S$ is an easy lambda term, the extension by $S=\lambda x.M$, with fresh $x$, is consistent. In that extensionThe theory obtained by adding only $ST=M$ is a subtheory of this consistent extension, so it is consistent too. Thus $ST$ is easy for every closed $T$. The printed unrestricted wording requires this qualification because easiness was defined only for closed terms: if $T$ is open, $ST$ need not be closed. With easiness extended to equations between open terms, the same argument works for arbitrary $T$.

If an easy closed $S$ were a solvable lambda term, some finite sequence of closed arguments would give $S{P}_1\cdots P_k{\equiv }_{\beta }I$. Repeated application of the result above would make this term easy. Easiness is invariant under beta equivalence, so $I$ would be easy, contrary to the preceding calculation. Hence every easy term is unsolvable.

An easy lambda term must be equatable consistently with every closed lambda term. Choose the proposed equation $I=K$. Applying it to $I$ gives $I=\lambda y.I$. Applying that equation to an arbitrary closed $V$ gives $V=I$. In particular the two Church Booleans become equal; then applying them to arbitrary $A,B$ gives $A=B$. The extended equational theory is trivial, hence inconsistent in the required sense. Therefore $I$ is not easy.