Formal Power Series as Linear Operators
Introductionary reference:
So why not replace $\,t\,$ by the differential operator $\,d/dx\,$ everywhere it occurs? Next reference:
Therefore we have in general and especially:
$$
e^{y\,d/dx} f(x) = f(x+y) \quad \Longrightarrow \quad e^{y\,d/dx} x^n = (x+y)^n
$$
Indeed the operator $1/(d/dx)$ is the inverse of differentiation, which is integration. So let's take a look at your perhaps not as obvious expression:
$$
\left(\frac{e^{y\,d/dx}-1}{d/dx}\right)x^n=\frac{1}{d/dx}\left(e^{y\,d/dx}-1\right)x^n\\
=\int\left[(x+y)^n - x^n\right] dx = \frac{(x+y)^{n+1}-x^{n+1}}{n+1} + C
$$
with $C$ an arbitrary constant. Evaluation is not ambiguous, because the operators $1/(d/dx)$ and $\left(e^{y\,d/dx}-1\right)$ do indeed commute. Which
means that we can do algebra with these operators as with common numbers. The same in, last but not least:
$$
\left(\frac{e^{y\,d/dx}-1-y\,d/dx}{(d/dx)^2}\right)x^n=\int\left\{\int\left[(x+y)^n-x^n\right]dx\right\}dx-\int y\,x^n\,dx
\\=\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + Cx + D
$$
with $C$ and $D$ arbitrary constants. This result is deviant from your last formula. It is noticed that:
$$
\frac{d^2}{dx^2} \left[\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} + C\,x + D\right]
= (x+y)^n - x^n - y\frac{dx^n}{dx}
$$
EDIT. I almost forgot to mention the stuff where it's all about: the formal power series
(Wikipedia).
$$
(y+x)^{n}=\sum _{k=0}^{n}{n \choose k}y^{n-k}x^{k} \quad \Longrightarrow \\
\frac{(x+y)^{n+2}-x^{n+2}}{(n+1)(n+2)}-\frac{y\,x^{n+1}}{n+1} =\\
\frac{1}{(n+1)(n+2)}\left[\sum _{k=0}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k} - x^{n+2} - (n+2)y\,x^{n+1}\right]=\\
\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}
+ {n+2 \choose 0}y^0 x^{n+2} - x^{n+2} + {n+2 \choose 1}y^1 x^{n+1} - (n+2)y\,x^{n+1} \right]\\
=\frac{1}{(n+1)(n+2)}\left[\sum _{k=2}^{n+2}{n+2 \choose k}y^{k}x^{n+2-k}\right]
$$
So here is your fix on the lower bound of the summation.