Let $\Phi(t)$ be $$\Phi(t)=\int\limits_t^\infty \kappa e^{-\int\limits_t^s g(\omega)d\omega} ds$$
such that $0<t<s$. If it holds that $\Phi(t)=\alpha$ where $\alpha$ is constant $\forall t$ can we prove the following by using Leibniz's rule for differentiation of a definite integral?
$$-\dot{\Phi}(t)+g(t)\Phi(t)-\kappa=0$$
$\underline{\text{Note:}}$ Recall that, by using the Leibniz integral rule then
$$\frac {\mathrm{d}}{\mathrm{d}x}\left(\int_{a(x)}^{b(x)}f(x, t) \,\mathrm{d}t\right)= f(x,b(x))\frac{\mathrm{d}}{\mathrm{d}x}b(x)- f(x, a(x))\dfrac{\mathrm{d}}{\mathrm{d}x}a(x)+ \displaystyle\int_{a(x)}^{b(x)}\dfrac{\partial f(x,t)}{\partial x} \,\mathrm{d}t$$
where $-\infty<a(t)$, $\beta(t)<\infty$ and the integrands are functions dependent on $x$.
Your assumption that $\Phi(t)=\text{constant}$ seems out of place. You just want to compute $\dot\Phi(t)$. As you've noted, when $\Phi(t) = \int_t^c h(t,s)\,ds$ (with $h$ and $\partial h/\partial t$ continuous, say), then $$\dot\Phi(t) = -h(t,t) + \int_t^c \frac{\partial h}{\partial t}ds.$$ In your case, $h$ looks forbidding, but we proceed: With $$h(t,s) = \kappa(s)e^{-\int_t^s g(\omega)d\omega}ds,$$ we have
\begin{align*} h(t,t) &= \kappa(t)e^0 = \kappa(t) \quad\text{and}\\ \frac{\partial h}{\partial t} &= \kappa(s) \left(e^{-\int_t^s g(\omega)d\omega}ds\right)\left(-(-g(t))\right) = \kappa(s)g(t)e^{-\int_t^s g(\omega)d\omega}ds. \end{align*}
Thus, $$\dot\Phi(t) = \kappa(t) + \int_t^\infty \kappa(s)g(t)e^{-\int_t^s g(\omega)d\omega}ds = \kappa(t) + g(t)\Phi(t),$$ as desired. (Apply differentiation under the integral sign with $\int_t^c$ and let $c\to\infty$. If $\Phi$ is well-defined in the first place, this will be fine.)