Are integral equations and differential equations allways equivalent, i.e. the set of solutions is equal? The following are my thoughts about it. Are they correct?
Let's define the function $x:\Bbb R_+\to\Bbb R$ by the integral equation $$x(t)=g(t)+\int_0^t f(x(s))\mathrm{d}s,\qquad x(0)=g(0)=a,$$ whereas $g\in\mathcal C^1(\Bbb R_+\to\Bbb R)$, $f\in\mathcal C(\Bbb R_+\to\Bbb R)$ and $a\in\Bbb R$.
We are looking for all solutions of $x$ in the space $\mathcal C(\Bbb R_+\to\Bbb R)$. Let's call this the integral equation problem (IEP).
Although not required in the conditions, from the equation already follows, that $x$ is in $\mathcal C^1(\Bbb R_+\to\Bbb R)$, since the right hand side is in $\mathcal C^1(\Bbb R_+\to\Bbb R)$ by the fundamental theorem of calculus and $g\in\mathcal C^1(\Bbb R_+\to\Bbb R)$, $f,x\in\mathcal C(\Bbb R_+\to\Bbb R)$. As $x$ is differentiable, we can differentiate the equation and get $$x'(t)=g'(t)+f(x(t)),\qquad x(0)=g(0)=a,$$ again with $g\in\mathcal C^1(\Bbb R_+\to\Bbb R)$, $f\in\mathcal C(\Bbb R_+\to\Bbb R)$ and $a\in\Bbb R$. The search for all solutions of $x$ in the space $\mathcal C^1(\Bbb R_+\to\Bbb R)$ we call the differential equation problem (DEP).
Simply by integrating we can transform the DEP back to the IEP. This shows that the IEP and the DEP are equivalent.
Are $g\in\mathcal{C}^1(\Bbb R_+\to\Bbb R)$, $f,x\in\mathcal C(\Bbb R_+\to\Bbb R)$ and $a\in\Bbb R$ the weakest assumptions we have to assume? I think so.
The reason why I think that the equivalence of the problems is helpful, is that for the DEP we have theorems like from Picard–Lindelöf, which can be used for the IEP then.
With the assumptions in the OP the two problems IEP and DEP are equivalent.
The continuity of $f$ guarantees the existence of a local solution, i.e., a solution defined in an interval $[0,T)$, which is not necessarily unique.
In order to obtain uniqueness, we need additional conditions on $f$. For example, local Lipschitz continuity or positivity.
In order to obtain a global existence, i.e., solution defined in $[0,\infty)$, we need further conditions on the growth rate of $f$. For example: $$ |f(x)|\le F(x) \quad\&\quad \int_0^\infty\frac{dx}{F(x)}=\infty. $$