First off, those comments (and the book in general) are actually about toposes, not general categories, and should be interpreted in that light. (This is something I've only come to realise recently, despite having read the book a couple of years ago.) Therefore one should look to categories of sheaves to make sense of the remarks.
And I am also puzzled by what exactly they meant by T=1. Did they mean "T= the category's terminal object"? Or "T= the category's separator object?" Or "T= a/the singleton object"? In Sets all these descriptions apply to 1, but it is not clear to me which of them is the basis for the authors' comments.
$1$ always means the terminal object in this book, if I remember correctly.
The notion of ‘singleton’ is very set-centric. It is possible to generalise this idea to toposes, but this is not what is meant here. Toposes in general do not have a separator object either; those that are separated by $1$ (and with $1$ not isomorphic to $0$) are called well-pointed.
Furthermore, it means that the authors are either (1) implying that the term injective can be meaningfully applied only to morphisms in categories that have a "1" object (with 1 = terminal/separator/singleton, depending on what the authors meant), or else (2) implying that every concrete category has one such object.
As already mentioned, not every category has a separator. Nor does every category have a terminal object. One can study the implied definition in any category with a terminal object, but it is not necessarily useful. For example, in categories where the terminal object is isomorphic to the initial object, e.g. $\mathbf{Grp}$ or $\mathbf{Ab}$, there will always be exactly one arrow $1 \to X$ for any object $X$. It is most profitable when the category is a topos. In the traditional case of categories of sheaves, an arrow $1 \to X$ is called a global section, so the comment is simply saying (for this case) that ‘a sheaf may not have enough global sections’, which is not surprising at all.
Indeed, consider the category $\mathbf{Set}^2$ consisting of pairs of sets and pairs of maps. The terminal object in this category is a pair of singleton sets. Consider the pair $(\emptyset, \{ * \})$. There is no arrow $(\{ * \}, \{ * \}) \to (\emptyset, \{ * \})$ because there is no map $\{ * \} \to \emptyset$. But clearly $(\emptyset, \{ * \})$ is not ‘empty’ (both in the intuitive sense and in the sense of not being an initial object). The book might call this an example of ‘not having enough global elements’. The way to rectify this is to consider ‘generalised elements’, which in the language of sheaves is akin to the notion of a local section. But notice that this category has a separating set, namely the pair of pairs $(\emptyset, \{ * \})$ and $(\{ * \}, \emptyset)$. It is tempting to take the coproduct of these two to try to make a single separating object, but we have already shown that doesn't work!
Now, to answer your first question. Note that monicity can be defined as a limit: an arrow $f : A \to B$ is monic if and only if the pullback of $A \xrightarrow{f} B \xleftarrow{f} A$ is the identity map. I will discuss only concrete categories here and ‘injective’ is reserved for maps of sets.
Necessary conditions. The above characterisation is useful because we can now understand preserving monics in terms of preserving limits. Thus, if the underlying set functor of a category preserves finite limits (or even just pullbacks), the underlying map of a monic arrow must be injective. This happens, for example, when the underlying set functor has a left adjoint (the ‘free object functor’). This means, for categories of where is a reasonable notion of ‘free object generated by one element’, the underlying map of a monomorphism must be injective.
Sufficient conditions. Similarly, if the underlying set functor reflects limits, then every arrow which has an injective underlying map must already be monic. This is typical in categories of algebraic objects and is true in particular for $\mathbf{Grp}$, $\mathbf{Ring}$, $\mathbf{Vect}$. Note, however, that the underlying set functor for $\mathbf{Top}$ does not reflect (or create) limits, so one has to search for other ‘reasons’ why a monic arrow in this category is the same thing as an injective continuous map. In fact, when the underlying set functor is faithful (i.e. injective on arrows), if the underlying map of an arrow is injective, then the arrow must be monic. The easiest way to see this is to use the fact that the functor $\mathrm{Hom}(C, -) : \mathbf{C} \to \mathbf{Set}$ preserves monics. (Exercise!)
A category with this property is called balanced, although I think this term is terrible.
In practice, the problem that arises is that the monomorphisms in some category correspond to our intuitive notion of the "injective" maps, but the epimorphisms frequently fail to correspond to our intuitive notion of the "surjective" maps; for example, in $\text{Ring}$, localizations like $\mathbb{Z} \to \mathbb{Q}$ are epimorphisms, and in Hausdorff spaces, any map with dense image is an epimorphism.
To get the "surjective" maps back usually requires working with a stronger notion of epimorphism, such as strong epimorphisms or effective epimorphisms. With any of these stronger notions that I'm aware of, a monomorphism which is an epimorphism in a stronger sense is an isomorphism. So the failure of monic epis to be isos can be thought of as the failure of epis to be some stronger notion of epi.
In an abelian category, more or less by definition, every epimorphism is effective, so we're fine. In general the condition that every epimorphism is effective is a form of the first isomorphism theorem, and I don't know of any abstract nonsense that easily guarantees it: it seems to me that abstract nonsense can't easily distinguish between the category of groups and the category of monoids, and the category of groups has the property that every epimorphism is effective while the category of monoids doesn't (it's again true that any localization is an epimorphism).
For concrete categories $U : C \to \text{Set}$, the second condition you describe (that isomorphisms lift) means that the forgetful functor $U$ is conservative. This is notably the case if $U$ is monadic (and in particular has a left adjoint). $U$ often preserves limits and hence monomorphisms, but it often fails to preserve epimorphisms and hence colimits. This is just a fact of life. (The forgetful functor from groups to sets notably preserves epimorphisms despite hardly preserving any colimits.)
You can get around all of this by proving things about some stronger notion of epimorphism rather than about epimorphisms.
Best Answer
In the category of rings with unity (with a $1$), morphisms are required to take the unity to the unity. In the case at hand, unless the morphism is trivial, the kernel will not be a ring with unity that embeds as a ring with unity.
Consider for example the case of $R=\mathbb{Z}\times\mathbb{Z}$, and the morphism $R\to\mathbb{Z}$ obtained by mapping $(a,b)$ to $a$. This is a ring-with-unit morphism.
The kernel of the map is the ideal $I=\{(0,b)\mid b\in\mathbb{Z}\}$. Now, as an abstract ring, $I$ is a ring with unity: the element $(0,1)$ is a unity for $I$. (In fact, $I$ is isomorphic as a ring-with-unity to $\mathbb{Z}$).
Thus, in this case, the kernel is in fact in the category: it is a ring with unity. However, the embedding $I\hookrightarrow R$ is not a morphism in the category of rings-with-unity, because the unit of $I$, $(0,1)$, is not mapped to the unit of $R$. Thus, the embedding is not a morphism in this category.
That’s Jacobson’s point: even if the kernel happens to, abstractly and by happenstance, be a ring with unity, you will almost never have that the embedding of the kernel into the ring is a morphism in the category. The slight error in your argument is that it is possible for the kernel to be a ring-with-unity, abstractly, even if it is not a subring-with-unity (which would require the unity of the ideal to be the same as the unity of the whole ring, in which case you are correct that the ideal would have to be the whole ring).
Note that this issue does not show up in the category of rings/rngs, where you do not require rings to have a unity and you do not require morphisms to map $1$ to $1$ even when the two rings do. In that category, the same proof shows that a morphism is monic if and only if it is one-to-one.