Difference of differentiation under integral sign between Lebesgue and Riemann

What you write is more or less true. First note that it is not enough to assume that $f(x,t)$ is differentiable at $x_0$ (for (almost) all $t$), we need that there is a neighorhood $(x_0 - \varepsilon, x_0 + \varepsilon) =:I$ such that $f(\cdot, t)$ is differentiable in $x$ for all $x \in I$ and (almost) all $t \in A$, where the "exceptional null set" must not depend on $x$.

The assumption $|\partial f/\partial x (x,t)| \leq g(x)$ only makes sense in this setting, otherwise we could only require $x_0$ instead of $x$ and the proof also needs this assumption (see below).

Now, let $h_n \to 0$ with $h_n \neq 0$ for all $n$ and w.l.o.g. with $x_0 + h_n \in I$ for all $n$. Let $$ F(x) := \int_A f(x,t) \, dt \text{ for } x \in I,\\ G(x) := \int_A \frac{\partial f}{\partial x}(x,t) \, dt \text{ for } x \in I. $$ Here, the integral defining $F$ is interpreted as an improper Riemann integral (it is possible that this integral does not exist as a Lebesgue integral), but the integral defining $G$ is interpreted as a Lebesgue integral. If $t \mapsto \frac{\partial f}{\partial x}$ is Riemann integrable over every compact subinterval of $A$ (for example, if it is continuous in $t$), then it can also be interpreted as an improper Riemann integral, at least if we assume that the dominating function $g$ is (improper) Riemann integrable.

Observe that the assumptions guarantee that $F,G$ are well-defined. Measurability of $t \mapsto \frac{\partial f}{\partial x}(x,t)$ for each $x \in I$ is implied by the pointwise convergence $$\frac{\partial f}{\partial x}(x,t) = \lim_n \frac{f(x+h_n, t) - f(x,t)}{h_n}, $$ which will also be employed below.

We now have \begin{eqnarray*} \left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| & = & \left|\int_{A}\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\,{\rm d}t\right|\\ & \leq & \int_{A}\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\right|\,{\rm d}t. \end{eqnarray*} Note that the integrands are measurable and that for (almost) all $t \in A$, the mean value theorem yields some $\xi_{n,t} \in I$ (even between $x_0$ and $x_0 +h_n$) such that $$ \left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}\right|=\left|\frac{\partial f}{\partial x}\left(\xi_{n,t},t\right)\right|\leq g\left(t\right). $$ This shows that the integrands in the integrals are Lebesgue-integrable and dominated by an integrable function. As the integrand in the last integral converges pointwise to $0$ for $n \to \infty$, the dominated convergence theorem yields $$ \left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| \to 0, $$ so that $F$ is differentiable (in $x_0$) with the expected derivative.