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.
I'd like to start with your first question about "almost kernel". The following construction can be brought to the case of "almost cokernel" without much modification.
Suppose $a:A\to M$ is an "almost kernel" and $k$ is the kernel map. By definition of $A$, $\varphi \circ a=0$ and there exists a map (not necessarily unique) $i:\text{Ker }\varphi \to A$ makes the following diagram commute:
$$\require{AMScd}
\begin{CD}
@. \text{Ker } \varphi @>i>> A @.\\
@. @| @VVaV \\
0 @>>> \text{Ker } \varphi @>k>> M @>\varphi>> N
\end{CD}$$
On the other hand, by the universal property of kernel there exists an unique map $r:A\to \text{Ker }\varphi$ makes the following diagram commute:
$$\require{AMScd}
\begin{CD}
A @>r>> \text{Ker } \varphi @.\\
@| @VVkV \\
A @>a>> M @>\varphi>> N
\end{CD}$$
We obtain another commutative diagram:
$$\require{AMScd}
\begin{CD}
@. \text{Ker } \varphi @>r\circ i>> \text{Ker }\varphi @.\\
@. @| @VVkV \\
0 @>>> \text{Ker } \varphi @>k>> M @>\varphi>> N
\end{CD}$$
Now by the universal property of kernel again $r\circ i=\text{id}_{\text{Ker}\varphi}$, since the identity also makes the above diagram commute. You can check that the existence of such $r,i$ is, in fact, the necessary and sufficient condition of an almost kernel. Ones can take $i$ to be the inclusion, then $r$ is called a retract of $A$ to $\text{Ker }\varphi$.
Such examples come from direct sums, since the composition of the projection from $M\oplus N$ onto $M$ and the obvious inclusion $M\to M\oplus N$ is the identity on $M$. In this case, you can take $\varphi: \mathbb{Z}\to 0$, then the kernel is $\text{id}:\mathbb{Z}\to \mathbb{Z}$ and an almost kernel is $\text{pr}_1: \mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}, (x,y)\mapsto x$. It is easy to see that $r=\text{pr}_1$ and a choice for $i:\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ is $x\mapsto (x,0)$. Another choice is $i':x\mapsto (x,x)$.
In the above argument, the use of universal property is essential. The main purpose of universal property is to obtain the uniqueness of the kernel (or cokernel). Suppose that there are two kernel $k:K\to M$ and $k':K'\to M$ of $\varphi:M\to N$, then a similar argument to the one we presented above show that there exists unique $i:K\to K'$ and $j:K'\to K$ such that $k'=k\circ j,k=k'\circ i$ and $i\circ j=\text{id}_{K'},j\circ i=\text{id}_K$. The last two equations say that $K$ and $K'$ are isomorphic, and the previous two equations say that the isomorphisms are natural.
The second advantage of the universal property which is not very clear at the moment is that universal property provides morphisms. This is the case with tensor product, and you will use this frequently when deals with diagrams ("diagram chasing" is the name of this method). Further, the notion of kernel and cokernel is important because they are the building blocks of abelian category. On an abelian category, we can do homological algebra, a ubiquitous computational tool of algebraists.
I think what I wrote above will give you some grasp about how universality work. From your question, I think you have the same problem as mine when I first learn about category theory. The definition of kernel suggests that there are many kernels, but in every application of universal property they treat them like there is only one kernel. From the categorical viewpoint, two objects defined by the same universal property are identical. When there is a kernel, its universal property allows one to only care about the unique factorization. This is quite troublesome at first, but it is indeed an advantage of category theory.
Best Answer
There is something crucial missing in your definition, which is that you need to additionally require that $\pi\varphi=0$. When you add that requirement, it becomes immediate that a cokernel can only every be the zero ring, since the map $0$ is not even a ring homomorphism unless the codomain is the zero ring.
(Note that if you applied your definition to $\mathbf{Ab}$, it would be highly non-unique for the same reason that you observed for rings and would not at all coincide with the usual meaning of cokernels.)