Consequences of not requiring ring homomorphisms to be unital?

This is just an amplification of Emerton and Pietro Majer's answers, but got a bit too big.

You can give a geometric interpretation of all not necessarily unital (NNU) $k$-algebras, for any ring k. The category of NNU $k$-algebras is equivalent to the category of augmented (unital) $k$-algebras, by which I mean $k$-algebras $A$ with a retraction $A\to k$ of the structure map, where the morphisms are the $k$-algebra maps that commute with the retractions. An augmented $k$-algebra $A$ corresponds to the NNU $k$-algebra which is the kernel of its augmentation map $A\to k$. Thus the category of NNU $k$-algebras is anti-equivalent to the category of pointed affine $k$-schemes, where the maps preserve the pointed structure. The forgetful functor from $k$-algebras to NNU $k$-algebras then has a geometric interpretation. It corresponds to the functor from affine $k$-schemes to pointed affine $k$-schemes that sends $X$ to the disjoint union of $X$ and $\mathrm{Spec}(k)$, where the second component is the distinguished point.

In particular, an NNU map $A\to B$ of $k$-algebras corresponds to a scheme map $$\mathrm{Spec}(B)\coprod \mathrm{Spec}(k) \to \mathrm{Spec}(A)\coprod \mathrm{Spec}(k)$$ which is the identity on the second component. The original map is unital if and only if the map takes the $\mathrm{Spec}(B)$ component to the $\mathrm{Spec}(A)$ component. But in general there could be a connected component of $\mathrm{Spec}(B)$ taken to the $\mathrm{Spec}(k)$ component. These are exactly the vaporized components in Emerton's answer.

Here's a little exercise. What does the NNU subring $n\mathbf{Z}$ of $\mathbf{Z}$ look like geometrically?

So the question about whether it's better to look at unital or NNU rings is (at least in the commutative case) the same as the question of whether it's better to looked a pointed or unpointed spaces, which also comes up in homotopy theory. I prefer the unital/unpointed approach (no doubt because of my education), but it's easy to translate back and forth between the two.

Let's suppose that our rings are commutative (which is the case that is immediately relevant to algebraic geometry).

If $\phi:A \to B$ is a (possibly non-unital) homomorphism, then $e := \phi(1_A)$ is an idempotent in $B$, and so we get a decomposition $B = eB \times (1-e)B,$ and the map $\phi$ factors as $A \to eB \to B,$ where the first map is unital, and the second map is simply the inclusion, which is the inclusion of a direct factor.

On Specs, we thus get the composite of the map Spec $eB \to $ Spec $A$, composed with the "map" (this is not necessarily an honest map of schemes, because it corresponds to the possibly non-unital map $e B \to B$) Spec $B \to$ Spec $eB$, which just vaporizes the open and closed subset Spec $(1-e)B$ of Spec $B$, and is the identity on the open and closed subset Spec $eB$.

So the upshot is that nothing much new happens in algebraic geometry, except that we allow maps which are only defined on some open and closed subset of a given scheme. Of course, this is a big except, because these are not honest maps at all (they are simply not defined on some part of their "domain"). There doesn't seem to be any reason to add them into the mix, which is surely one reason why this generalized notion of homomorphism is not used much in practice.

P.S. One could argue another way, beginning with geometry, and passing to algebra by remembering that rings are rings of functions. If we have a map $\phi:X \to Y$ of spaces (of some type, e.g. affine schemes, or anything else), then surely the constant function 1 on $Y$ will pull-back to the constant function 1 on $X$. Thus the induced homomorphism on rings of functions will have to be unital, and so one simply has no cause to consider non-unital homomorphisms in the geometric setting.

P.P.S. The argument in the first paragraph shows that allowing non-unital homomorphisms in the category of commutative rings is the same as adding, in addition to unital homomorphisms, homomorphisms of the form $B_1 \to B_1\times B_2,$ given by $b_1\mapsto (b_1,0),$ for any pair of commutative rings $B_1$ and $B_2$. So it's not really a very exciting change from the purely algebraic point of view either.

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.