$f(a)+f(-b)=0$ implies $f(-a)+f(b)=0$. Then $f$ is an odd function?

If $f(a)+f(-b)=0$ for some $(a,b)$, then, $f(-a)+f(b)=0$.

Also, for all $a$, there exists $b$ such that $f(a)+f(-b)=0$.

$f:\mathbb R\to\mathbb R$ is continuous.

Based on the above conditions, can we prove or disprove that $f$ is an odd function: $f(x)=-f(-x)$?

I think we cannot prove this. Being odd seems to be a sufficient but not necessary. Yet I failed to find any counterexamples against the claim.

Update: Thanks to user WhatsUp, even functions $f(x)=f(-x)$ can also satisfy these conditions.

User Kavi suggests that if we add $f(x)>0$ for all $x>0$, then $f$ is odd. (please correct me if I am wrong) I will make a new question based on his intuition.

A quite simple counterexample is $f(x) = \cos(x)$.

$$\forall a ~ \exists b: f(a)+f(-b)=0 $$ Choose $a=0$. There exists $b\in \mathbb{R}$ such that $f(0)+f(-b)=0$

From the first condition, $f(0)+f(b)=0$ $$ \therefore \exists b \in \mathbb{R} :f(b) = f(-b)$$ Thus, your function is certainly not odd.