It is readily checked that any affine function satisfies the condition in the statement. To prove the converse, set $y = x$ to write
$$\int_{0}^{2x} f(t) \, dt = x(f(2x) + f(0)). \quad (*)$$
Since $f$ is integrable, $(*)$ shows that the function $g: \mathbb{R} \to \mathbb{R}$, $g(x) = x(f(2x) + f(0))$, is continuous, so $f$ is continuous on $\mathbb{R}^* = \mathbb{R} \setminus \{0\}$. Continuity of $f$ on $\mathbb{R}^*$ and $(*)$ then show $g$ differentiable on $\mathbb{R}^*$, so $f$ is differentiable on $\mathbb{R}^*$.
Differentiate $(*)$ to get $2f(2x) = f(2x) + f(0) + 2x f'(2x)$; that is, $x f'(x) - f(x) + f(0) = 0$ for all $x \neq 0$. Consequently,
$$\left( \frac{f(x) - f(0)}{x} \right)' = 0 \quad \text{for all } x \neq 0,$$
so
$$f(x) = \begin{cases} a x + f(0), & x < 0, \\ b x + f(0), & x > 0. \end{cases}$$
Set now $x = 0$ and $y = 1$ in the relation in the statement to write
$$\begin{aligned} f(1) + f(-1) &= \int_{-1}^{1} f(t) \, dt = \int_{-1}^{0} (a t + f(0)) \, dt + \int_{0}^{1} (b t + f(0)) \, dt \\ &= \frac{1}{2}(b-a) + 2f(0). \end{aligned}$$
Plug $f(1) = b + f(0)$ and $f(-1) = -a + f(0)$ into the above relation to get $b - a + 2f(0) = \frac{1}{2}(b-a) + 2f(0)$, so $a = b$ and $f(x) = a x + f(0)$ for all real numbers $x$.

Alternative Solution:
For convenience, consider the integrable function $g: \mathbb{R} \to \mathbb{R}$, $g(x) = f(x) - f(0)$, along with the continuous function $G: \mathbb{R} \to \mathbb{R}$, $G(x) = \int_{0}^{x} g(t) \, dt$. Clearly, $g$ and $G$ both vanish at the origin, and it is readily checked that
$$G(x+y) - G(x-y) = y(g(x+y) + g(x-y)), \quad \text{for all real numbers } x \text{ and } y. \quad (1)$$
Let $x = y = t/2$ in (1), and recall that $g$ and $G$ both vanish at the origin, to infer that
$$2G(t) = t g(t), \quad \text{for all real numbers } t. \tag{2}$$
Next, let $x = 0$ in (1) and use (2) to infer that $G$ is an even function: $G(-y) = G(y)$, for all real numbers $y$. With reference again to (2), it follows that $g$ is odd: $g(-y) = -g(y)$, for all real numbers $y$. It is therefore sufficient to focus on the restrictions $g_+$ and $G_+$ of $g$ and $G$, respectively, to the ray of positive real numbers.
By (2), continuity of $G_+$ implies that of $g_+$ which in turn implies differentiability of $G_+$ and $G'_+ = g_+$.
Hence $2G_+(t) = t G'_+(t)$ for all real $t > 0$, showing that the function $t \mapsto t^{-2} G_+(t)$, $t > 0$, is constant; that is, $G_+(t) = \alpha t^2$ for some real constant $\alpha$, so $g_+(t) = a t$, where $a = 2\alpha$.
Consequently, $g(t) = a t$ for all real numbers $t$, and $f(t) = a t + f(0)$.