What is an example of a non-zero "ring pseudo-homomorphism"? By "pseudo ring homomorphism", I mean a map $f: R \to S$ satisfying all ring homomorphism axioms except for $f(1_R)=f(1_S)$.
Even if we let this last condition drop, there are only two ring pseudo-homomorphisms from $\Bbb Z$ to $\Bbb Z[\omega]$ where $\omega$ is any root of unity, for example. They are the identity homomorphism and the "$0$ pseudo-homomorphism".
I couldn't find a non-zero example. It is easy to conclude that $0=f(a)(1_S-f(1_R))$, so I know that the counterexample will have to be some $S$ with zero-divisors.

Every commutative example takes the following form. Suppose $f : R \to S$ is a non-unital ring homomorphism between two commutative rings. Then $f(1_R)$ is some idempotent $m \in S$, as Jendrik Stelzner remarks. $mS$ is a "non-unital" subring of $S$ (it's a subring except that its unit is $m$, not $1_S$), and $f$ is a ring homomorphism in the ordinary sense to this subring. Moreover, $S$ decomposes as a product of rings
$$S \cong mS \times (1 - m)S.$$
So darij's comment essentially exhausts all examples. 
Geometrically such a morphism corresponds to a "partially defined" morphism $\text{Spec } S \to \text{Spec } R$ of affine schemes, where "partially defined" means defined on some union of connected components. There are analogous statements one can make about non-unital C*-algebra homomorphisms between commutative C*-algebras. 
