Question: As defined in many modern algebra books, a homomorphism of unital rings must preserve the unit elements: $f(1\_R)=1\_S$. But there has been a minority who do not require this, one prominent example being Herstein in *Topics in Algebra*.

What are some of the most striking consequences of not requiring ring homomorphisms to be unital? For example, what aspects of algebraic geometry would need to be reworked if we no longer required it? What interesting theorems or techniques arise in the not-necessarily-unital theory which do not apply (or are degenerate) for unital homomorphisms?

Answer: 
Sometimes you work with some non-unital rings and homomorphisms between them, but the rings happen to be unital. This happens in the context of AF-algebras: This is a $C^{\ast}$-algebra, which is a directed colimit of finite-dimensional $C^{\ast}$-algebras. It turns out that every finite-dimensional $C^{\ast}$-algebra is a direct product of matrix algebras, thus unital. In order to classify AF-algebras, you have to study the transition maps, which are non-unital ring homomorphisms between unital rings. So this happens quite naturally. It turns out that there is a way of visualizing these direct systems via so called Bratteli diagrams. The simplest example is the colimit of the imbeddings $M\_n \to M\_{n+1}, A \mapsto diag(A,0)$, which is the stabilized matrix algebra $M\_\infty$. If you are interested in this, check out Davidson's "$C^{\ast}$-algebras by example". The commutative case is equivalent to the study of non-unital boolean rings, which may also be seen as colimits of finite, thus unital boolean rings.

Now you asked about algebraic geometry. I think one main issue is that if $\phi : A \to B$ is a non-unital homomorphism between unital rings, then there is no reason why prime ideals of $B$ pull back to prime ideals of $A$. Indeed, the inclusion $A \to A \times B$ pulls back $A \times \mathfrak{p}$ to $A$. So you cannot define an associated map of spectra $Spec(B) \to Spec(A)$. As Emerton already pointed out, anyway you get a map $Spec(Be) \to Spec(A)$, where $e=\phi(1)$.

This can be seen functorial in the following sense: Let $C$ be the category of (perhaps non-unital) rings together with a choosen idempotent element; homomorphisms should respect this idempotent element. Then there is a functor $C \to Ring$, which sends $(A,e) \mapsto Ae$, which can be composed with $Spec : Ring \to LRS$. If you plug in $(A,1)$ and $(B,e)$, you get the map above.
If we permit rings to have **1** = **0**, then the trivial ring {**0**} becomes the zero object of the category.

For comparison, if we required morphisms to keep **1** (but still permitted for **1** = **0**), then {**0**} would be the terminal object only – any object has (unique) morphism to {**0**}, and **1** maps to **1** because **1** = **0** in the trivial ring. But there would be no morphisms from {**0**} to anything else since **1** ≠ **0** in any non-trivial ring.