The taylor approximation of $\frac{1}{x}$ does not exist centered at $x = 0$, as the function is not continuous there. However, there does exist a formula:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n\\ \Rightarrow \frac{1}{1 - (1 - x)} = \sum_{n=0}^\infty (1 - x)^n = \frac{1}{x} $$
However, this function is not of a definite parity (i.e.) $f(-x) \neq -f(x)$ and $f(x) \neq f(-x)$. This is a characteristic of the original function $\frac{1}{x}$.
Is there a polynomial approximation, that maybe looks something like the below sketch?
This functions is odd and seems plausible to make. I just cannot find it online. Is there any function like what I drew? Again, I am not looking for a taylor approximation, as it won't exist. But perhaps an ad-hoc solution has been made?
You can use the Bernstein polynomials (after some rescaling) to approximate the functions $$g_n(x)=\frac{\text{sgn}(x)}{|x|+1/n}$$ in the interval $[-1,1]$ since we have $g_n(x)$ converges pointwise to $\frac{1}{x}$ for $x\ne 0$.
That gives $$f_n(x)=\sum_{k=0}^n g_n(2\tfrac{k}{n}-1)b_{k,n}(\tfrac{x+1}{2})$$
Here's a graph of some of the approximations: