A singular Gronwall inequality

No, for $$ f(t)=t, $$ we have $f(0)=0$ and $$ \int_0^t s^{-1}f(s)\,ds=t=f(t),\quad\forall t>0, $$ so (your constant is $1$, and you actually have equality) $$ f(t)\leq 1\int_0^t s^{-1}f(s)\,ds,\quad\forall t>0. $$

If we set $g(t)=\frac{f(t)}{t}$ and $G(t)=\int_{0}^{t}g(u)\,du$, we have the inequality: $$ t\cdot g(t)\leq C\cdot G(t)\tag{1} $$ or: $$ \frac{g(t)}{G(t)}\leq \frac{C}{t}\tag{2} $$ hence integrating both sides over $[\varepsilon,x]$ we have: $$ \log(G(t))-\log(G(\varepsilon))\leq C\left(\log(t)-\log(\varepsilon)\right)\tag{3}$$ or: $$ G(t)\leq G(\varepsilon)\left(\frac{t}{\varepsilon}\right)^C\tag{4} $$ but as shown by mickep, by choosing $g\equiv 1\neq 0$ and $C=1$ we have that $(1)$ is fulfilled.