Suppose that we have $f: \mathbb{R}\rightarrow\mathbb{R}$ is Riemann integrable and
$$ \int_0^1 f(x+\theta(y-x))d\theta \leq \frac{f(x)+f(y)}{2} \ \ \ (\ast) $$
for all $x,y\in\mathbb{R}$. Is this sufficient to conclude that $f$ is convex. I know that converse holds. I have also noticed that $(\ast)$ is equivalent to $$ \int_0^1 f(x+\theta(x-y))d\theta \leq \int_0^1 f(x)+\theta(f(y)-f(x))d\theta. $$
It is then obvious, from monotonicity of integration, if $f$ is either strictly non-negative or strictly non-positive we have $f$ is convex. But what if we know nothing about the sign of $f$?
Let $|f| \leq M$ and $g=M+1+f$. Observe that $g$ satisfies the hypothesis and $g$ is strictly positive. From your observation $g$ is convex and this implies $f=g-M-1$ is convex.