(Some obvious properties of $E$; too long for a comment, though).

The holomorphic extension of $E$ to $\mathbb{C} \setminus (-\infty, 0]$ (in fact, to the entire Riemann surface of the complex logarithm) is given by $$E(x) = \int_0^\infty \frac{\exp(t \log x)}{\Gamma(t+1)}\, dt,$$ where $\log$ denotes the principal branch of the complex logarithm. This follows from a standard application of Morera's theorem, involving Fubini's theorem, the estimate $$|\exp(t \log x)| = \exp(t \log |x|) = |x|^t \le a^t + b^t$$ when $a \le |x| \le b$, and integrability of $(a^t + b^t) / \Gamma(t + 1)$ over $(0, \infty)$.

In particular, for $x > 0$ we have $\log(-x + 0 i) = \log x + i \pi$, and hence
$$ E(-x+0i) = \int_0^\infty e^{i \pi t} \frac{x^t}{\Gamma(t+1)} \, dt $$
is not real-valued in any neighbourhood of $0$. Thus, there is no real-analytic extension of $E$ to $(-\epsilon, \infty)$ (as already follows from Carlo Beenakker's comment).

By dominated convergence theorem, the integral can be differentiated under the integral sign, so
$$E^{(n)}(x) = \int_0^\infty \frac{x^{t - n}}{\Gamma(t+1 - n)}\, dt = \int_{-n}^\infty \frac{x^t}{\Gamma(t+1)}\, dt .$$
This does not seem to lead to any interesting differential equation.

Since $E'(x) > 0$, clearly $E$ is increasing on $(0, \infty)$, with $E(0) = 0$ and $E(\infty) = \infty$.

One can easily find the Laplace transform of $E(x)$: when $\operatorname{Re} \xi > 1$, we have
$$ \int_0^\infty e^{-\xi x} E(x) dx = \int_0^\infty \frac{1}{\xi^{t + 1}} \, dt = \frac{1}{\xi \log \xi} . $$

*(EDIT: This was meant to be an extended comment only, but since it has received a number of upvotes, let me add further remarks, inspired by Nemo's answer.)*

The Laplace transform $\mathcal{L} E$ of $E$ has a simple pole at $\xi = 1$ with residue $1$, and a branch cut along $(-\infty, 0]$. Since it decays (barely) sufficiently fast at infinity, one can (carefully) write the usual inversion formula and then deform the contour of integration to the Hankel contour to find that
$$ E(x) = e^x - \frac{1}{\pi} \int_0^\infty e^{-t x} \operatorname{Im} (\mathcal{L} E(-t + 0i)) dt .$$
This leads to the formula given in Nemo's answer: since $$\mathcal{L} E(-t + 0i) = -\frac{1}{t \log(-t + 0 i)} = -\frac{1}{t (\log t + i \pi)} \, ,$$ we obtain
$$ E(x) = e^x - \int_0^\infty \frac{e^{-t x}}{t (\pi^2 + \log^2 t)} \, dt .$$
As a consequence, $e^x - E(x)$ is completely monotone, and
$$ E(x) = e^x - \frac{1 + o(1)}{\log x} $$
as $x \to \infty$. Further terms can be obtained in a similar way.

The function $E(x)$ itself is the Mellin transform of $1 / \Gamma(t + 1)$. Thus, $1 / \Gamma(t + 1)$ can be written as the inverse Mellin transform:
$$ \frac{1}{\Gamma(t + 1)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-1 - x} E(x) dx , $$
or, equivalently,
$$ \frac{1}{\Gamma(t)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-x} E(x) dx . $$
The definition of $E(x)$ looks a little bit like Mellin–Barnes integral, but the contour is wrong.

Finally, the (fractional) integral of $E(x)$ of order $\alpha$ is given by
$$ I_\alpha E(x) = \frac{1}{\Gamma(\alpha)} \int_0^x E(t) (x - t)^{\alpha - 1} dt = \int_0^\infty \frac{x^{t + \alpha}}{\Gamma(t + 1 + \alpha)} \, dt , $$
and so
$$ I_\alpha E(x) = \int_\alpha^\infty \frac{x^t}{\Gamma(t + 1)} \, dt . $$
This agrees with the expression for the derivatives of $E$ (which correspond to negative integer $\alpha$).