If $g=g(t)$ is integrable on $\mathbb{R}$ and satisfies an integral equation $$g(t) = a + b\int_s^tg(u) du,\quad t\geq s,$$ then we have a solution $ g(t) = a \exp{\{b(t-s)\}}.$
More generally, for an increasing and continuous function $A=A(t)$ on $\mathbb{R}$, let us consider an integral equation $$ g(t) = a + b\int_s^tg(u) dA(u),\quad t\geq s$$ where $a$ and $b$ are constants, and the integral is a Lebesgue-Stieltjes integral, and $g$ is integrable w.r.t. $A$ on every interval $[s,t]$.
Then, how can I solve the equation above ? I expect it would have a solution $$g(t) = a \exp{\{b(A(t)-A(s))\}}. $$
Suppose $A$ is $C^1$. Then the equation is $$ g(t) = a + b\int_s^tg(u)\,A'(u)\,du,\quad t\geq s. $$ Taking the derivative with respect to $t$ be obtain the differential equation $$ g'(t)=b\,A'(t)\,g(t),\quad g(s)=a. $$ Write it as $g'/g=b\,A'$ and integrate to obtain $g=C\,e^{bA}$ for some constant $C$, whose value can be found from the condition $g(s)=a$, resulting in $c=a\,e^{-bA(s)}$. finally, we get as you expected $$ g(t)=a\,e^{b(A(t)-A(s)}. $$