Clearly, every constant function satisfies the required conditions. Conversely, write the condition in the statement in the equivalent form
$$\int_{x}^{x+y} f(t) \, dt \le \int_{0}^{y} f(t) \, dt$$
to infer that $\int_{0}^{y} f(t+x) \, dt \le \int_{0}^{y} f(t) \, dt$ for all non-negative $x$ and $y$.
On the other hand, since $f$ is increasing, $f(t+x) \ge f(t)$ for all $t$ in the closed interval $[0, y]$ and all $x \ge 0$, so $\int_{0}^{y} f(t+x) \, dt \ge \int_{0}^{y} f(t) \, dt$.
Consequently, $\int_{0}^{y} f(t+x) \, dt = \int_{0}^{y} f(t) \, dt$ for all non-negative $x$ and $y$. Continuity of $f$ forces $f(x+y) = f(y)$ for all non-negative $x$ and $y$. In particular, $f(x) = f(0)$ for all $x \ge 0$, so $f$ is constant.