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.
 A: Because $f(1_R) = f(1_R^2) = f(1_R)^2$ be know that $1_R$ must be mapped to an idempotent element of $S$. We can use this to construct some counterexamples:
Take for example a commutative ring $R$ and a not-necessarily commutative or unital (but associative) $R$-algebra $A$. For any idempotent $e \in A$ the map $R \to A, r \mapsto re$ does the trick. This map is only unital if $A$ is unital and $e = 1_A$.
One examples of this is a ring (i.e. $\mathbb{Z}$-algebra) $S$ with an idempotent $e \in S$. (If $ne = 0$ for some $n \geq 2$ then this also results a pseudo ring homomorphism $\mathbb{Z}/n \to S$. Matt Samuel’s answer is an example of this with $S = \mathbb{Z}/6$, $e = 3$ and $n = 2$.)
If $k$ is a field then the matrix algebra $\mathrm{M}_n(k)$ gives us lots of idempotents (unless $k$ is finite and $n$ is small, or $n = 1$), as there is a bijection between the idempotents of $\mathrm{M}_n(k)$ and the direct sum decompositions $k^n = U \oplus V$ (where we regard $U \oplus V$ and $V \oplus U$ as two different decompositions).
Another class of examples arises by taking a product of unital rings $\prod_{i \in I} R_i$ and considering the idempotent $e_j = (\delta_{ij})_{i \in I} \in \prod_{i \in I} R_i$; we can use this to construct the pseudo ring homomorphism $R_j \to \prod_{i \in I} R_i$, $r \mapsto r e_j$, which is only unital if $|I| = 1$.
A: 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. 
A: Try the homomorphism
$$\mathbb Z_2\to \mathbb Z_6$$
given by 
$$[1]_2\mapsto [3]_6$$
A: The (conceptually) simplest example I know is $f:\Bbb Z\to\Bbb Z\times\Bbb Z$ defined by $f(n)=(n,0)$. The ring identity of $\Bbb Z\times\Bbb Z$ is $(1,1)$ but $1$ is instead mapped to $(1,0)$. The same example shows that a subset of a ring that is a ring is not necessarily a subring if you don't require it to contain the ring unit.
