There is a relationship between these two concepts. Let the complex function $f(z)$ be analytic on and inside a simple closed curve $C$ in the complex plane. Then Cauchy's integral formula states that for any point $z$ inside $C$ we have
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=2\pi j f(z)\tag{1}$$
For $z$ outside $C$, it follows from Cauchy's theorem that
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=0\tag{2}$$
Furthermore, for $z$ on the curve $C$ the following holds:
$$\oint_C\frac{f(\zeta)}{\zeta-z}d\zeta=\pi j f(z)\tag{3}$$
A proof of Eq. $(3)$ can be found in any text on complex analysis.
Let's now choose the curve $C$ as the real line from $-R$ to $R$ and the semi-circle $Re^{j\phi}$, $\phi\in[0,\pi]$, such that $C$ is closed. For $R\to\infty$, the curve $C$ "encloses" the upper half-plane. If $f(z)$ is analytic in the upper half-plane, and if it decays to zero sufficiently fast such that the contribution of the integral along the semi-circle with radius $R$ vanishes as $R\to\infty$, Eq. $(3)$ becomes
$$\int_{-\infty}^{\infty}\frac{f(\zeta)}{\zeta -x}d\zeta=\pi j f(x)\tag{4}$$
Note that the integral in $(4)$ is along the real line. With $z=x+jy$ and $f(z)=f_R(x,y)+jf_I(x,y)$, Eq. $(4)$ can be written as
$$\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)+jf_I(\zeta,0)}{\zeta -x}d\zeta=\pi j \big[f_R(x,0)+jf_I(x,0)\big]\tag{5}$$
Splitting Eq. $(5)$ into real and imaginary parts yields
\begin{align} f_I(x,0) & = -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_R(\zeta,0)}{\zeta-x}d\zeta &&= \mathscr{H}\big\{f_R(x,0)\big\} \tag{6}\\ f_R(x,0) & = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f_I(\zeta,0)}{\zeta-x}d\zeta &&= -\mathscr{H}\big\{f_I(x,0)\big\}\tag{7} \end{align}
where $\mathscr{H}\{\cdot\}$ denotes the Hilbert transform.
Equations $(6)$ and $(7)$ are identical to Eqs $(1)$ and $(2)$ in the question with $x_R(t)=f_R(t,0)$ and $x_I(t)=f_I(t,0)$. Consequently, the analytic function $f(z)$ equals an analytic signal on the real line.
Example 1:
Given the analytic signal $x(t)=e^{j\omega_0t}$, $\omega_0>0$, the corresponding analytic function is given by $f(z)=e^{j\omega_0z}$. Note that $f(z)$ satisfies all properties required for Eqs $(4)-(7)$ to be valid: it is analytic everywhere, and it decays rapidly for $|z|\to\infty$ and $\textrm{Im}\{z\}>0$ (i.e., in the upper half-plane).
Example 2:
We can also use functions with the required properties (analytic and decaying sufficiently fast in the upper half plane) to generate Hilbert transform pairs, or, equivalently, analytic signals. Let
$$f(z)=\frac{e^{j\omega_0 z}-1}{j\pi z},\qquad\omega_0>0\tag{8}$$
The function $f(z)$ is analytic everywhere (if we define $f(0)=\omega_0/\pi$), and it decays rapidly in the upper half plane $\textrm{Im}\{z\}>0$. Consequently, by evaluating $f(z)$ on the real line, we obtain an analytic signal and a Hilbert transform pair. For notational convenience, let's use $t$ to denote the real part of $z$. The corresponding analytic signal is given by
\begin{align} x(t) &= \frac{e^{j\omega_0 t}-1}{j\pi t} \\ & = \frac{\cos(\omega_0t)+j\sin(\omega_0t) - 1}{j\pi t} \\ & = \frac{\sin(\omega_0t)}{\pi t} + j\frac{1 - \cos(\omega_0t)}{\pi t} \tag{9} \end{align}
It has been shown in the answers to this question that the real and imaginary parts of $(9)$ indeed form a Hilbert transform pair. From the same answers, the Fourier transform of $x(t)$ can be easily derived:
$$X(j\omega)=\begin{cases}2,&0<\omega<\omega_0\\0,&\textrm{otherwise}\end{cases}\tag{10}$$
showing (again) that $x(t)$ is an analytic signal.
In a completely analogous manner we can show the relation between the causality of a time domain function and the analyticity of the corresponding transfer function. If $h(t)$ is the impulse response of a causal LTI system, i.e., if $h(t)=0$ for $t<0$ is satisfied, it is well-known that the real and imaginary parts of its Fourier transform (the frequency response) are related by the Hilbert transform. Let $H(j\omega)$ be the Fourier transform of $h(t)$ with real and imaginary parts $H_R(\omega)$ and $H_I(\omega)$, respectively:
$$\mathscr{F}\big\{h(t)\big\} = H(j\omega) = H_R(\omega) + jH_I(\omega) $$
Then the following Hilbert transform relations hold:
\begin{align} H_R(\omega) &= \mathscr{H}\big\{H_I(\omega)\big\}\tag{11} \\ H_I(\omega) &= -\mathscr{H}\big\{H_R(\omega)\big\}\tag{12} \end{align}
Now let's evoke Eq. $(3)$, but this time we choose the curve $C$ as the straight line connecting the points $z=jR$ and $z=-jR$ on the imaginary axis, and the semi-circle $Re^{j\phi}$, $\phi\in[-\pi/2,\pi/2]$, connecting those two points via the right half-plane. For a stable system, the transfer function $H(s)$ will decay sufficiently fast in the right half-plane, such that the contribution to the total integral $(3)$ of the integral over the semi-circle vanishes for $R\to\infty$, and we're left with an integral over the imaginary axis:
\begin{align} \lim_{R\to\infty}\oint_C\frac{H(\zeta)}{\zeta-s}d\zeta &= \int_{j\infty}^{-j\infty}\frac{H(\zeta)}{\zeta-s}d\zeta \\ & = \int_{-\infty}^{\infty}\frac{H(j\Omega)}{s-j\Omega}jd\Omega\tag{13} \end{align}
For $s$ on $C$, i.e., on the imaginary axis, we have $s=j\omega$ and, using $(13)$, Eq. $(3)$ becomes
$$\int_{-\infty}^{\infty}\frac{H(j\Omega)}{\omega-\Omega}d\Omega = \pi j H(j\omega)\tag{14}$$
Splitting $(14)$ into real and imaginary parts yields
\begin{align} H_R(\omega) &= \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{H_I(\Omega)}{\omega-\Omega}d\Omega \\ H_I(\omega) &= -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{H_R(\Omega)}{\omega-\Omega}d\Omega \end{align}
which are just the Hilbert transform relations $(11)$ and $(12)$. I.e., the Hilbert transform relations are equivalent to Eq. $(14)$, which is a consequence of Eq. $(3)$ IF $H(s)$ is analytic in the right half-plane and if it decays sufficiently fast in the right half-plane. In sum, causality (and stability) of an LTI system cause its transfer function $H(s)$ to be analytic in the right half-plane. This is equivalent to $H(s)$ satisfying the Hilbert transform relations $(11)$ and $(12)$ on the imaginary axis.
Depending on the field, the Hilbert transform relations $(11)$ and $(12)$ are also referred to as Kramers-Kronig relations or as dispersion relations (e.g., here and here).
Example 3:
Given the causal and stable impulse response
$$h(t)=e^{-t}u(t)$$
we know from the above that the corresponding transfer function
$$H(s)=\frac{1}{1+s}$$
must be analytic and decaying in the right half-plane, both of which is straightforward to verify. Furthermore, $H(s)$ must satisfy the Hilbert transform relations $(11)$ and $(12)$ on the imaginary axis. Splitting $H(j\omega)$ into real and imaginary parts gives
$$\mathscr{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}$$
which is a well-known Hilbert transform pair (see this answer or this table).
