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: 
Of course, if we assume that the ring have a unit, then there is absolutely no reason not to assume that homomorphisms preserve it (or are there books that do that?)

My impression is that there has been a change in the last 40 years or so; algebra texts written in the '60's or earlier mostly did not require ring to be unital, while, say, from the '70's on most of them did. Of course, rings without unit are still very much present in functional analysis (for example, Banach algebras are not assumed to be unital, as many of the standard examples are not). The first edition of Herstein's book is from 1964.

I don't know how much of this is due to Grothendieck's influence. Certainly algebraists now are less conversant with functional analysis than they used to be, due to an inevitable increase in specialization. In the kind of mathematics I practice, units are a fact of life. The only advantage I can see in not assuming the existence of a unit is that ideals become rings, and one can apply theorems about rings; but this is minor, and is more than offset by the conceptual disadvantages of working with the "wrong" category.
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.