Some Motivation
Recall that a (covariant) functor $F \colon \mathscr{C} \to \mathsf{Set}$ is representable by an object $A$ of $\mathscr{C}$ if and only if there exists a universal element $a \in F(A)$.
This means that there exists for every other object $X$ of $\mathscr{C}$ and every element $x \in F(X)$ a unique morphism $f \colon A \to X$ in $\mathscr{C}$ with $F(f)(a) = x$.
The natural bijection $\operatorname{Hom}_{\mathscr{C}}(A, -) \to F$ is then given by $f \mapsto F(f)(a)$.
A useful example to keep in mind in the functor $F \colon \mathsf{Ring} \to \mathsf{Set}$ given by $F(R) = R^n$.
This functor is represented by the ring $A = \mathbb{Z}\langle t_1, \dotsc, t_n \rangle$, the (non-commutative) polynomial ring in the variables $t_1, \dotsc, t_n$.
The (usual) universal element $a \in F(A)$ is given by $a = (t_1, \dotsc, t_n)$.
And indeed, that this is a universal element means precisely that there exists for every other ring $R$ and every element $x \in F(R)$ with $x = (x_1, \dotsc, x_n)$ a unique ring homomorphism $f \colon A \to R$ with $F(f)(a) = x$, i.e. with $f(t_i) = x_i$ for every $i = 1, \dotsc, n$.
And this is precisely how the ring $A$ represents the functor $F$.
The Problem
We show more generally that for every number of elements $n \geq 2$ the functor
\begin{align*}
F
\colon
\mathsf{Ring}
&\to
\mathsf{Set},
\\
R
&\mapsto
\{
(x_1, \dotsc, x_n) \in R^n
\mid
x_1 R + \dotsb + x_n R = R
\}
\end{align*}
is not representable.
Assume otherwise that the functor $F$ is representable by a ring $A$ and let $(a_1, \dotsc, a_n) \in F(A)$ be the universal element (corresponding to some choice of isomorphism $F \cong \operatorname{Hom}_{\mathsf{Ring}}(A,-)$).
We will consider some auxilary functors which are representable:
We can consider for every index $i = 1, \dotsc, n$ the functor
\begin{align*}
E_i
\colon
\mathsf{Ring}
&\to
\mathsf{Set},
\\
R
&\mapsto
\{
(x_1, \dotsc, x_n) \in R^n
\mid
\text{$x_i$ is a unit in $R$}
\} \,.
\end{align*}
This functor is representable by the ring $U_i := \mathbb{Z}\langle t_1, \dotsc, t_n, t_i^{-1} \rangle$.
We can also consider for any two indices $i$, $j$ with $1 \leq i \neq j \leq n$ the functor
\begin{align*}
E_i
\colon
\mathsf{Ring}
&\to
\mathsf{Set},
\\
R
&\mapsto
\{
(x_1, \dotsc, x_n) \in R^n
\mid
\text{$x_i$ and $x_j$ are units in $R$}
\} \,.
\end{align*}
This functor is representable by the ring $U_{ij} := \mathbb{Z}\langle t_1, \dotsc, t_n, t_i^{-1}, t_j^{-1} \rangle$.
We fix for the rest of this argumentation two such indices $i$, $j$.
(This is where we use that $n \geq 2$.)
We have inclusions of functors as follows:
These inclusions correspond to ring homomorphism between their representing objects:
The inclusion $E_i \subseteq E_{ij}$ correspond to the canonical ring homomorphisms $U_i \to U_{ij}$, and the inclusion $E_i \subseteq F$ correspond to the ring homomorphisms $f_i \colon A \to U_i$ with $f(a_k) = t_k$ for every index $k = 1, \dotsc, n$.
(Such a homomorphism exists because it is the unique homorphism $f \colon A \to U_i$ with $F(f)( (a_1, \dotsc, a_n) ) = (t_1, \dotsc, t_n)$.
So here we use that $(a_1, \dotsc, a_n)$ is a universal element and that $(t_1, \dotsc, t_n)$ is an element of $F(U_i)$.)
Similar for $U_j$ instead of $U_i$.
The commutativity of the above diagram gives (by the faithfulness of the Yoneda embedding) the commutativity of the corresponding diagram:
This means that $f_i(a) = f_j(a)$ for every element $a \in A$.
It follows that $f_i$ and $f_j$ restrict to the same ring homomorphism $f \colon A \to U_i \cap U_j$, with the intersection $U_i \cap U_j$ taken in $U_{ij}$.
This intersection is precisely the usual polynomial ring $U := \mathbb{Z}\langle t_1, \dotsc, t_n \rangle$.
We have seen in the above explicit description of the homomorphism $f_i$ (and $f_j$) that the homomorphism $f \colon A \to U$ is given by $f(a_k) = t_k$ for every index $k = 1, \dotsc, n$.
The existence of such a homorphism means that $(t_1, \dotsc, t_n) \in F(U)$ bescause the ring $A$ represents the functor $F$ via the universal element $(a_1, \dotsc, a_n) \in F(A)$.
But this would means that $U = t_1 U + \dotsb + t_n U$, which is not the case.
We hence see that such a representing object $A$ cannot exist.
Best Answer
I assume that by "ring" you mean "commutative ring." Suppose $\text{Nil}$ is represented by some commutative ring $N$. Then $\text{id}_N \in \text{Hom}(N, N) \cong \text{Nil}(N)$ must be the "universal nilpotent" $n \in N$: that is, it is a nilpotent with the property that it maps to every other nilpotent in every other commutative ring $R$ under a (unique) homomorphism $N \to R$. (Every representable functor works this way: see universal element. This is $a$ in your work.)
But there can't be a universal nilpotent, because any nilpotent element must be nilpotent of some particular degree $k$. And if $n^k = 0$ then $n$ can't map to nilpotents of degree larger than $k$. For a bunch of variations on this argument see this blog post. In your work you attempt to map the universal nilpotent to something which is not a nilpotent at all.
Alternatively although similarly, you can argue that $\text{Nil}$ doesn't preserve infinite limits. (A representable covariant functor preserves all limits.) In fact it already fails to preserve infinite products, as follows: consider the product
$$R = \prod_{k \in \mathbb{N}} \mathbb{Z}[x]/x^k.$$
Then each $x \in \mathbb{Z}[x]/x^k$ is nilpotent but the product element $\prod x$ is not.