One can introduce a constant $\chi$ following all numerical properties and two additional operations $\delta$ and $\circ$, like the following:
$\delta a=0$ if a is a standard number
$\delta \chi=1$
$\delta (xz)=z\delta x+x\delta z$
If
$x=a+a_1\chi+a_2\chi^2+a_3\chi^3+...$
and
$z=b+b_1\chi+b_2\chi^2+b_3\chi^3+...$
Then $x\circ z=a+a_1z+a_2z^2+a_3z^3+...$
so that $\chi\circ z=z$ and $z\circ\chi=z$ for any $z$
This system seems to me to possess all usual properties of real/complex numbers, and form a field. Yet it is often claimed that the usual complex numbers are the only system of (hyper-)complex numbers that forms a field. Why?
Also the proposed system will make operators and functionals just ordinary numerical functions.
The reason this does not form a field is quite simply the lack of multiplicative inverses. (it is clearly a ring, namely $\mathbb{C}[\chi]$).
Under standard multiplication $\chi$ has no inverse as $k\chi \notin \mathbb{C}$ if $k\in \mathbb{C}$ and $\chi *\chi = \chi^2$.
So assuming you wish to use composition of functions as your multiplication, we have $2+\chi$ having no inverse:
Suppose $\alpha = (2+\chi)^{-1}$ then $1 = (2+\chi) \circ \alpha = 2+ \alpha $ from the definition of $\circ$ hence $\alpha = -1$ but $\alpha \circ (2+\chi) = -1 \circ (2+\chi) = -1$ which is not the identity.