If $\int_{0}^{1}f=\int_{0}^{1}g=1$, then there exists an interval $I \subset [0,1]$ such that $\int_{I}f=\int_{I} g =\frac{1}{2}$

I know you said that $f$ and $g$ aren't necessarily non-negative, but here's a proof in the case that $f\geq 0$ (and $g$ is arbitrary). Perhaps you can use the ideas for the general case?

Let $I = [0,1]$ for ease of notation. Consider the map $\phi:I^2\rightarrow \mathbb{R}^2$ with $\phi(x,y) = \left( \int_x^y f(t)-g(t)\; dt, \int_x^y f(t)\; dt - \frac{1}{2}\right)$. Note that $\phi(x,y) = 0$ iff $I =[x,y]$ or $[y,x]$ solves your problem.

Assume for a contradiction that $\phi(x,y)\neq (0,0)$ for all $x,y\in I$. Then define $\psi:I^2\rightarrow S^1$ by $\psi(x,y) = \frac{\phi(x,y)}{|\phi(x,y)|}$. This is well defined because $\phi(x,y)\neq (0,0)$ by assumption.

Now, consider $\psi|_{\partial I^2}:\partial I^2\rightarrow S^1$ (where I'm imagining I've chosen a homeomorphism $\partial I^2\cong S^1$). This map has degree $0$ since extends to $I^2$.

On the other hand, we will now show the degree is non-zero. This contradiction will establish that $\phi(x,y) = (0,0)$ for some $(x,y)$, giving you your desired interval.

First, a direct computation establishes that $\psi(1,0)=\psi(0,0) =\psi(1,1)= (0,-1)$ and $\psi(0,1) = (0,1)$.

Let $\ell_i$ for $i=1$ to $4$ denote the four sides of $I^2$, with $\ell_1$ connecting $(0,0)$ to $(1,0)$, and rest proceeding counterclockwise.

Along $\ell_1$, we have $\phi(x,0) = \left( \int_x^0 f(t)-g(t)\; dt, \int_x^0 f(t)\; dt - \frac{1}{2}\right).$ Because $f\geq 0$, the second component of $\phi(x,0)$ is always negative.

In particular $\psi|_{\ell_1}$ has image in the third and 4th quadrant. So, the closed curve $\psi|_{\ell_1}$ can be homotoped to a constant without the homotopy every hitting $(0,0)$.

Along $\ell_2$, we have $\phi(1,y) = \left(\int_1^y f(t)-g(t)\; dt, \int_1^y f(t)\; dt - \frac{1}{2}\right)$. Again, the second coordinate is always negative since $f\geq 0$. So, as before, the closed curve $\psi|_{\ell_2}$ can be homotoped to a constant without the homotopy ever hitting $(0,0)$.

Along $\ell_3$, we have $\phi(x,1) = \left(\int_x^1 f(t)-g(t)\; dt, \int_x^1 f(t)\; dt - \frac{1}{2}\right)$. This time, $\psi|_{\ell_3}$ is not a closed curve. But the second coordinate is decreasing in $x$. This means that $\psi|_{\ell_3}$ is homotopy rel end points to an arc connecting $(0,1)$ and $(0,-1)$.

Along $\ell_4$, we have $\phi(0,y) = \left( \int_0^y f(t)-g(t)\; dt, \int_0^y f(t)\; dt - \frac{1}{2}\right)$. As before, since $f\geq 0$, the second coordinate is monotonically increasing, so $\psi|_{\ell_4}$ is homotopy rel end points to an arc connecting $(0,1)$ to $(0,-1)$.

To finish the proof, we show that the two arcs coming from $\ell_3$ and $\ell_4$ pass by different sides of $(0,0)$, proving that $\psi_{\partial I^2}$ has degree $\pm 1$.

So, fix an $x_0$ for which $\psi|_{\ell_3}(x_0,1)$ has second coordinate $0$, that is, for which $\int_{x_0}^1 f(t)\; dt = \frac{1}{2}$. Since $\psi(x_0,1)\neq (0,0)$ (since $\psi(x,1)\in S^1$), the first coordinate of $\psi(x_0,1)$, $\int_{x_0}^1 f(t) - g(t)\; dt$, must be non-zero, so is positive or negative. Let's assume it's positive, the other case being analogous.

Note that if $x_1$ is any other value for which $\psi(x_1,1)$ has vanishing second coordinate, then the sign of the first coordinate must be positive as well. Indeed, along the interval $[x_1,x_0]$ or $[x_0,x_1]$, $\psi(x, 1)$ has vanishing second coordinate (since the second coordinate is monotonic in $x$), and since the first coordinate of $\psi(x, 1)$ is nonvanishing on $[x_1,x_0]$, it must have the same sign on both end points.

Now, consider the point $(0,x_0)\in \ell_4$. The second coordinate of $\psi(0,x_0)$ is $\int_0^{x_0} f(t)\; dt - \frac{1}{2}$. But since $\int_{x_0}^1 f(t)\; dt = \frac{1}{2}$ and $\int_0^1 f(t)\; dt = 1$, it follows that $\int_0^{x_0} f(t)\;dt = \frac{1}{2}$ as well.

Then the first coordinate of $\psi(0,x_0)$ is $\int_0^{x_0} f(t)-g(t)\; dt$. We already know that $\int_{x_0}^1 f(t) - g(t)\; dt$ is positive, and that $\int_0^1 f(t) - g(t)\; dt = 0$. It follows that $\int_0^{x_0} f(t) - g(t)\; dt$ is negative. Thus, the arc coming from $\ell_4$ goes on the opposite side of $(0,0)$ as the arc coming from $\ell_3$. Thus, $\psi|_{\partial I^2}$ has degree $\pm 1$, giving our contradiction.