I think your calculations are correct.
It follows from the condition
$$
x
=
\sum_{(x)} x_{(1)} \varepsilon( x_{(2)} )
=
x \varepsilon(x) - y \varepsilon(y)
$$
that $\varepsilon(x) = 1$, and then also $\varepsilon(y) = 0$.
The antipode does then have to satisfy
$$
1
=
\varepsilon(x)
=
\sum_{(x)} x_{(1)} S( x_{(2)} )
=
x S(x) - y S(y)\,.
$$
But the right hand side of this equation has no constant part.
This equality is thus not possible.
I think there are two ways of fixing this example.
First way
Let $T$ be a two-dimensional vector space with basis $c$, $s$.
One can make $T$ into a coalgebra by setting
$$
\Delta(c)
=
c \otimes c - s \otimes s \,
\quad
\Delta(s)
=
c \otimes s + s \otimes c \,.
$$
The counit is given by
$$
\varepsilon(c) = 1 \,,
\quad
\varepsilon(s) = 0 \,.
$$
This is (sometimes) called the trigonometric coalgebra.
One should think about the basis elements $c$ and $s$ as the functions $\cos$ and $\sin$, and the comultiplication as a version of the addition theorems for cosine and sine.
This heuristic becomes justified by the following example.
Second way, preparation
Let $X$ be a set and let $\mathbb{k}$ be a field.
Let $F(X)$ be the set of functions from $X$ to $\mathbb{k}$.
This is a algebra via the pointwise addition, scalar multiplication and multiplication of functions, i.e. via
\begin{align*}
(f_1 + f_2)(x) &= f_1(x) + f_2(x) \, \\
(\lambda f)(x) &= \lambda f(x) \,, \\
(f_1 f_2)(x) &= f_1(x) f_2(x) \,.
\end{align*}
We have a map
\begin{align*}
\Phi
\colon
F(X) \otimes F(X)
&\to
F(X \times X) \,,
\\
f_1 \otimes f_2
&\mapsto
[ (x_1, x_2) \mapsto f_1(x_1) f_2(x_2) ] \,.
\end{align*}
This map is an injective homomorphism of algebras.
If the set $X$ is finite then by comparing dimensions we find that the homomorphism $\Phi$ is an isomorphism.
Let now $G$ be a finite group.
The multiplication of $G$ is a map
$$
m \colon G \times G \to G \,.
$$
This map induces a homomorphism of algebras
$$
m^*
\colon
F(G)
\to
F(G \times G) \,,
\quad
f
\mapsto
f \circ m \,.
$$
More explicitely,
$$
m^*(f)(g_1, g_2)
=
f(g_1 g_2) \,.
$$
Under the isomorphism $\Phi^{-1}$ we can regard the homomorphism $m^*$ as a homomorphism of algebras
$$
\Delta
\colon
F(G)
\to
F(G) \otimes F(G) \,.
$$
More explicitely, in terms of Sweedler notation,
$$
f( g_1 g_2 )
=
\sum_{(f)} f_{(1)}( g_1 ) f_{(2)}( g_2 ) \,.
$$
Let us denote the unit element of $G$ by $e$.
The inclusion map $\{ e \} \to G$ induces an algebra homomorphism $F(G) \to F(\{e\})$.
We may identify $F(\{e\})$ with the ground field $\mathbb{k}$ via the evaluation map $f \mapsto f(e)$.
We then get overall a homomorphism of algebras
$$
\varepsilon
\colon
F(G)
\to
\mathbb{k} \,,
\quad
f
\mapsto
f(e) \,.
$$
Lastly, the inversion map
$$
i
\colon
G
\to
G \,,
\quad
g
\mapsto
g^{-1}
$$
induces an algebra homomorphism
$$
S
:=
i^*
\colon
F(G)
\to
F(G) \,,
\quad
f
\mapsto
f \circ i \,.
$$
More explicitely,
$$
S(f)(g)
=
f( g^{-1} ).
$$
The algebra $F(G)$ together with the homomorphisms $\Delta$, $\varepsilon$ and $S$ is a Hopf algebra.
For an inifinite group $G$ this construction does in general not work because the injective homomorphism $\Phi$ won’t be an isomorphism.
Second way
We would now like to take $\mathbb{k} = \mathbb{R}$ and for the finite group $G$ the additive group $(\mathbb{R}, +)$.
But this group isn’t finite.
We can try to circumvent this problem by considering instead of $F(\mathbb{R})$ only a suitable subalgebra $H$, such that $m^*(H)$ is contained in $\Phi( H \otimes H )$, and $S(H) \subseteq H$.
We can then make $H$ into a Hopf algebra in the same way as above.
For our example we choose $H$ as the subalgebra of $F(\mathbb{R})$ which is generated by the two functions $\cos$ and $\sin$.
We have
\begin{align*}
m^*(\cos)(x, y)
&=
\cos(x + y)
\\
&=
\cos(x) \cos(y) - \sin(x) \sin(y)
\\
&=
\Phi( \cos \otimes \cos - \sin \otimes \sin )(x, y)
\end{align*}
and
\begin{align*}
m^*(\sin)(x,y)
&=
\sin(x + y)
\\
&=
\cos(x)\sin(y) + \sin(x)\cos(y)
\\
&=
\Phi( \cos \otimes \sin + \sin \otimes \cos )(x,y) \,,
\end{align*}
as well as
$$
S(\cos)(x)
=
\cos(-x)
=
\cos(x)
$$
and
$$
S(\sin)(x)
=
\sin(-x)
=
-\sin(x) \,.
$$
We now find that the algebra $H$ is a Hopf algebra with comultiplication
\begin{align*}
\Delta(\cos) &= \cos \otimes \cos - \sin \otimes \sin \,, \\
\Delta(\sin) &= \cos \otimes \sin + \sin \otimes \cos \,,
\end{align*}
counit
$$
\varepsilon( \cos ) = \cos(0) = 1 \,,
\quad
\varepsilon( \sin ) = \sin(0) = 0 \,,
$$
and antipode
$$
S(\cos) = \cos \,,
\quad
S(\sin) = -\sin \,.
$$
The functions $\sin$ and $\cos$ satisfy the relation
$$
\cos^2 + \sin^2 = 1 \,.
$$
We can follow from this that the algebra $H$ is isomorphic to the algebra
$$
H' = \mathbb{R}[c, s] / (c^2 + s^2 - 1) \,,
$$
such that $c$ corresponds to $\cos$ and $s$ corresponds to $s$.
The algebra $H'$ does now become a Hopf algebra with comultiplication
$$
\Delta(c) = c \otimes c - s \otimes s \,,
\quad
\Delta(s) = c \otimes s - s \otimes c \,,
$$
counit
$$
\varepsilon(c) = 1 \,,
\quad
\varepsilon(s) = 0 \,,
$$
and antipode
$$
S(c) = c \,,
\quad
S(s) = -s \,.
$$
We note this in the quotient $\mathbb{R}[c, s] / (c^2 + s^2 - 1)$ our original problem with the antipode doesn’t occur because
$$
\sum_{(c)} c_{(1)} S( c_{(2)} )
=
c S(c) - s S(s)
=
c^2 + s^2
=
1
=
\varepsilon(c) \,.
$$
