Identity related to $\sum_{k=0}^{n}\frac{x^k}{\binom{n}{k}}$

Your result is a kind of generalization of Newton's binomial identity with binomial coefficients replaced by their inverses.

You will find in page 2 of this reference by Toufik Mansour, University of Haifa) the more general expression :

$$\sum_{k=0}^{n}\frac{a^kb^{n-k}}{\binom{n}{k}}=\frac{n+1}{(a+b)\left(\tfrac{1}{a}+\tfrac{1}{b}\right)^{n+1}}\sum_{k=1}^{n+1}\frac{(a^k+b^k)\left(\tfrac{1}{a}+\tfrac{1}{b}\right)^{k}}{k}\tag{1}$$

with its proof. Nice expression...

It suffices now to take $a=x$ and $b=1$...

I have to admit that the proof is long... and uses generating functions.

Remark : If in the following identity for Beta integrals (see here):

$$B(x,y)=\int_{t=0}^{t=1}t^{x-1}(1-t)^{y-1}dt=\frac{(x-1)!(y-1)!}{(x+y-1)!}\tag{2}$$

one takes $x-1=k$ and $y-1=n-k$, we deduce that :

$$\frac{1}{\binom{n}{k}}=(n+1)\int_{t=0}^{t=1}t^k(1-t)^{n-k}dt\tag{3}$$

(besides, (3) is mentionned in the referenced article).

It is likely that a (simpler) proof of your identity can be deduced from (3) by multiplying it by $x^k$ and summing from $k=0$ to $k=n$.