First, rewrite your equation as $$ \frac{dy(t)}{dt}=b+f(t)(y(t))^n, \qquad \qquad \qquad (*) $$ where $f(t)=1/a(t)$.
For $n=1$ you have a linear inhomogeneous ODE which is easily solved.
For $n=2$ you get a special case of the so-called general Riccati equation $$ \frac{dy(t)}{dt}=b+f(t)(y(t))^2, $$ solving which is equivalent to solving a second-order linear ODE. Indeed, introduce a new independent variable $\tau(t)=\int a(t) dt$ you end up with the "standard" Riccati equation $$ \frac{dy(\tau)}{d\tau}=h(\tau)+(y(\tau))^2, $$ where the function $h$ is defined so that $h(\tau(t))=b a(t)$. and putting $y(\tau)=-(1/z(\tau))dz(\tau)/d\tau$ yields $$ d^2z(\tau)/d\tau^2+h(\tau)z(\tau)=0. $$ However, this linear equation in general is not necessarily solvable by quadratures.
For $n=3$ (*) is a special case of the Abel differential equation of the first kind, see e.g. here and references therein for details.
For $n\neq 1,2,3$ your equation in full generality is probably not solvable by quadratures but some of its special cases may be, see e.g. the book (in German)
E. KAMKE: Differentialgleichungen, Lösungen und Lösungsmethoden, Band I, Leipzig, 1951,
and this list.