Let $g:\mathbb R \to [0,\infty)$ be a function satisfying $g(0)=0$, which is strictly increasing on $[0,\infty)$, and strictly decreasing on $(-\infty,0]$.
Suppose that $$g(tx+(1-t)y) \le tg(x) + (1-t)g(y)$$ holds for every $t \in [0,1]$ and every $x,y$ such that $x+y \le 0$.
Is $g$ convex on all $\mathbb R$?
The assumption clearly implies that $g|_{(-\infty,0]}$ is convex.
Such a $g$ need not be convex. Consider $$g(x) = \begin{cases}\quad - x &\text{if } x \leqslant 0, \\ \log (1+x) &\text{if } x \geqslant 0. \end{cases}$$
Since $g(-x) \geqslant g(x)$ for all $x \geqslant 0$, this function satisfies the hypothesis, but on $[0,\infty)$ it is strictly concave.