The continuous Taylor series; are they just Taylor series? I first posed this question when I was a first year student. I came up with some ad hoc arguments as to why the result is true (a bit of numerical experimentation), but never had a proof. I forgot about it until the other day, and thought it was still an interesting question. First, some motivation.
Take the expression
$$\sum_{n=0}^\infty \frac{x^n}{n!}$$
which equals its own derivative, that much is straight forward. The $x^n$ gets sent to $n x^{n-1}$ and the factorial kills the $n$, and the same expression emerges. 
We can do the same thing with the integral.
$$\int_{-\infty}^\infty \frac{x^y}{\Gamma(y+1)}\,dy$$
formally satisfies being its own derivative--albeit the expression does not converge (which is a shame). But if you act like it converges, it should converge to $e^x$ because it's its own derivative.
This led me to considering a more suitable case, where the integral actually converges, and considering if the two functions agree. For example,
$$F(x) = \int_{-\infty}^{\infty}e^{-y^2} \frac{x^y}{\Gamma(y+1)}\,dy$$
defines a $C^{\infty}$ function in $x$ for $x \in \mathbb{R}^+$. This function satisfies $F'(x) = e^{-1}F(\frac{x}{e^2})$.  
If we take
$$G(x) = \sum_{n=0}^\infty e^{-n^2}\frac{x^n}{n!}$$
then equally so, $G'(x) = e^{-1}G(\frac{x}{e^2})$. 

Does $G = F$?

I thought of this in a more general setting though. I wondered if it happened for $e^{-y^2}$, does it work for other functions?
It may be useful to explain the ad hoc arguments I had when I was younger. These arguments were a summation of identities which are satisfied by both expressions, which seems implausible if they weren't equal in some sense. I'll explain them in the obtuse language I used when I was a first year, as it is the best way I can think of framing it. Let $\mathcal{I}$ and $\mathcal{S}$ denote the following operators,
$$\mathcal{I} \{f\} (x) = \int_{-\infty}^\infty f(y)\frac{x^y}{\Gamma(y+1)}\,dy$$
$$\mathcal{S} \{f\} (x) = \sum_{n=0}^\infty f(n) \frac{x^n}{n!}$$
Also, let $Qf(x) = xf(x)$ and $Tf(x) = f(x+1)$. Then $\mathcal{S}$ and $\mathcal{I}$ behave identically under actions of the two operators. The following similar identities present evidence as to why these two operators may be equivalent in a good enough scenario.
$$\mathcal{I}\{Tf\}(x) = \frac{d}{dx}\mathcal{I} \{f\} (x)$$
$$\mathcal{S} \{Tf\} (x) = \frac{d}{dx}\mathcal{S} \{f\} (x)$$
$$\mathcal{I}\{Qf\}(x) = Q\mathcal{I}\{Tf\}(x) = Q\frac{d}{dx}\mathcal{I} \{f\} (x)$$
$$\mathcal{S} \{Qf\} (x) = Q\mathcal{S} \{Tf\} (x) = Q\frac{d}{dx}\mathcal{S} \{f\} (x)$$
From these we also get
$$Q \mathcal{I}\{f\}(x) = \mathcal{I}\{QT^{-1}f\}(x)$$
$$Q \mathcal{S}\{f\}(x) = \mathcal{S}\{QT^{-1}f\}(x)$$
From all this the question is simple,

Are these two operators ever equivalent, if so when?

Or phrased more poetically

Is the continuous Taylor series no different than the discrete Taylor series?

 A: No. In fact I will show the map from $1$-periodic signed measures $\mu$ to $$F_\mu(x) = \int_{-\infty}^\infty \mu(y) e^{-y^2}\frac{x^y}{\Gamma(y+1)}dy$$ is injective for "reasonable" signed measures $\mu$.
Indeed, consider what happens to the following expression as $y_0$ goes to $\infty$ while the residue of $y_0$ modulo $1$ remains fixed: 
$$ \Gamma(y_0+1)  e^{-y_0^2 - y_0 \operatorname{dlog} \Gamma(y_0+1) } F_{\mu} ( e^{ 2 y_0 + \operatorname{dlog}  \Gamma(y_0+1)} ) $$
$$ = \int_{-\infty}^{\infty} \mu(y) e^{-y^2 + 2 y y_0 - y_0^2} \frac{ \Gamma(y_0+1) e^{ y \operatorname{dlog} \Gamma(y_0+1)- y_0 \operatorname{dlog} \Gamma(y_0+1)}}{\Gamma(y+1)} dy$$
$$ = \int_{\infty}^{\infty} \mu(y+y_0) e^{-y^2} \frac{ \Gamma(y_0+1) e^{ y \operatorname{dlog} \Gamma(y_0+1)}}{\Gamma(y+y_0+1)} dy$$
Because $\Gamma(y+1)$ is log-convex and its slope is roughly $\log y$, the exponential-linear approximation to the slope gets better as $y_0$ goes to $\infty$, and so this
$$ \approx \int_{-\infty}^{\infty} \mu(y+y_0) e^{-y^2} dy $$
uniquely determines the convolution of $\mu$ with the Gaussian. Now because none of the Fourier transform coefficients of the Gaussian vanish, we can extract all the Fourier series coefficients of $\mu$ from this, which under mild analytic conditions extracts $\mu$. 
In particular, for your example $F(x)$ we see that $F(e^{2 y_0 + \operatorname{dlog} \Gamma(y_0+1)})$ is asymptotic to a constant times $\frac{   e^{y_0^2 + y_0 \operatorname{dlog} \Gamma(y_0+1) }}{ \Gamma(y_0+1)}$ while this is not true for your Taylor series $G(x)$.
A: So I think I found a way of proving $F \neq G$ using distribution theory. I'm not the best at distribution theory, it's definitely not my field, but I think I'm doing something right. This is as much rigor as I can produce in distribution theory, but I think I'm on the right track. I don't have a referential brain for distribution theory, I'm going off what I can remember, I apologize if I'm not explicitly stating the theorems I'm using. This idea works generally as to why the continuous Taylor series can never equal the discrete Taylor series, but I'll be a bit cavalier. Maybe this argument works in more general cases, but one would need to know more about distribution theory to produce such a result, sadly I don't.
Let's just use $e^{-y^2}$ to be explanatory. If we take
$$F(a,x) = \int_{-\infty}^\infty e^{-y^2} e^{-2\pi a y i} \frac{x^y}{\Gamma(y+1)}\,dy$$
Then we can look at the equivalency in a Fourier sense.
Take the Poisson summation
$$\sum_{n=-\infty}^\infty F(n,x) = \sum _{n=-\infty}^\infty e^{-n^2} \frac{x^n}{n!} = \sum _{n=0}^\infty e^{-n^2} \frac{x^n}{n!}$$
which follows because $\frac{1}{n!} = 0$ for $n<0$. Now I'm not certain about the following interchange, but I'm fairly certain it should be allowed in good enough scenarios (especially with $e^{-y^2}$).
$$\sum_{n=-\infty}^\infty F(n,x) = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^{-y^2} e^{-2\pi i n y} \frac{x^y}{\Gamma(y+1)}\,dy = \int_{-\infty}^\infty \chi(y)e^{-y^2} \frac{x^y}{\Gamma(y+1)}\,dy$$
where $\chi$ is the dirac comb. This is kind of nothing new to what Christian Remling said, except I'm phrasing it distributionally, which is to our advantage as we add the Fourier transform. As least, if I've not mistaken myself. The above expression can be thought of as an inner product. If the discrete and continuous Taylor series equal each other, then $(\chi, h) = (1,h)$ or $(\chi -1, h) =0$, where
$$h(y) = e^{-y^2}\frac{x^y}{\Gamma(y+1)}$$
I'm kind of out of my element right here, if I'm wrong, definitely correct me. Let $\mathcal{F}$ denote the Fourier Transform. Let $\delta$ denote the dirac $\delta$ function
$$\mathcal{F}\{\chi -1\} = \sum_{n=-\infty}^\infty e^{2\pi i n y} - \delta = \chi - \delta$$
satisfies
$$(\mathcal{F}\{\chi-1\},h) = \sum_{n=1}^\infty e^{-n^2}\frac{x^n}{n!} \neq 0$$
Notice we remove the $n=0$ term. This produces the final result. It is straight forward from here! We just use the unitary properties of $\mathcal{F}$ under this inner product.
$$(\chi-1,\mathcal{F}^{-1}h) \neq 0$$
Apply $\mathcal{F}$ again, using the unitary property again, and through the exact same argument
$$(\chi-1,h) = (\mathcal{F}\{\chi -1\}, \mathcal{F}^{-1}h) = (\chi - \delta, \mathcal{F}^{-1} h)\neq 0 $$
therefore
$$(\chi-1,h) \neq 0$$
The continuous Taylor series cannot equal the discrete Taylor series. Personally, I'm ashamed to ask, but for what functions does this actually work. I know this argument must work in a distributional sense, but for what functions does this actually work when written in a more classical sense?
