The continuous Taylor series; are they just Taylor series?

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 $G(x)$.