Infinite Integration by Parts
This is a restatement of the proof of Taylor's theorem using integration by parts. We have $$g(x)=g(a)+(x-a)g'(a)+\dots+ \frac{(x-a)^{n}}{n!}g^{(n)}(a)+\frac{1}{n!}\int_{a}^{x}g^{(n+1)} (t) (x-t) ^{n}\,dt\tag{1}$$ Now we put $g(x) = \int_{a}^{x} f(t) \, dt$ to get $$\int_{a} ^{x} f(t) \, dt=(x-a)f(a)+\dots+\frac{(x-a)^{n}}{n!}f^{(n-1)}(a)+\frac{1}{n!}\int_{a}^{x}f^{(n)}(t)(x-t)^{n}\,dt\tag{2}$$ Putting $x=0$ in the above equation we get your formula with $a$ in place of $x$.
It is no wonder that you used integration by parts to arrive at your formula because the equation $(1)$ above is also obtained via integration by parts and is presented in a routine manner in many textbooks and also on Wikipedia. But since you arrived at this sincerely by your own efforts hats off to you! +1 for your question.
Proof and Some Intuition
The OP claims that $$\int_{0}^{x} f(t) \, dt= \sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} f^{(n-1)}(x)$$ We can rewrite the RHS to get
$$\int_{0}^{x} f(t) \, dt=\sum_{n=0}^\infty \frac{x^{n+1}}{n!} (-1)^n f^{(n)}(x)$$ Differentiating both sides and assuming $f(0)=0$ we get $$f(x)=\sum_{n=0}^\infty\frac{(-1)^n x^n \left[x f^{(n + 1)}(x) + (n + 1) f^{(n)}(x)\right]}{(n + 1)!}$$ $$f(x)=\sum_{n=1}^\infty\frac{(-1)^{n-1} x^nf^{(n)}(x)}{n!}+\sum_{n=0}^\infty\frac{(-1)^n x^n f^{(n)}(x)}{n!}$$
Edit:
@MattWatkins pointed out that the partial sums cancel out. Now that I see this, it is obvious; notice that the series are the same but with flipped sign except for the first term of the second series, which never gets canceled. We thus see that our series will telescope, and what we are left with is $$f(x) = \frac{(-1)^0 x^0 f^{(0)}(x)}{0!} = f(x)$$ which proves the OP's result and provides a bit of the motivation for the series. I would still be interested in seeing if this could be applied though.
Edit 2
We can look at the OP's series and turn it into some sort of Taylor Series $$\sum_{n=1}^\infty \frac{x^n}{n!} (-1)^{n - 1} f^{(n-1)}(x)$$ $$= -\sum_{n=1}^\infty \frac{(0-x)^n}{n!} f^{(n-1)}(x)$$ $$= -\sum_{n=0}^\infty \frac{(0-x)^{n+1}}{(n+1)!} f^{n}(x)$$ Notice that this immediately looks like the integral of a Taylor Series, providing further motivation and hinting that there is an intuitive justification for the OP's series.