You are right, this is not true in general.
Proof by contradiction. If your statement is true, then we have
$$
E_{xc}[\rho] = \sum_i E_{xc}[\rho_i] \tag{1}
$$
whenever
$$
\rho = \sum_i \rho_i \tag{2}
$$
Now, every density can be expressed as an infinite sum of delta functions. Since the sum is on all points in $\mathbb{R}^3$, we may prefer to write it as an integral instead:
$$
\rho = \int \rho_{\mathbf{r}'} d\mathbf{r}', \quad \rho_{\mathbf{r}'}(\mathbf{r}) = \rho(\mathbf{r}')\delta(\mathbf{r}-\mathbf{r}') \tag{3}
$$
This means that
$$
E_{xc}[\rho] = \int E_{xc}[\rho_{\mathbf{r}'}] d\mathbf{r}' \tag{4}
$$
Since $E_{xc}[\rho_{\mathbf{r}'}]$ only depends on the value of $\rho$ at the point $\mathbf{r}'$, Eq. (4) means that $E_{xc}[\rho]$ is an LDA functional:
$$
E_{xc}[\rho] = \int \epsilon_{xc}(\rho(\mathbf{r})) d\mathbf{r}, \quad \epsilon_{xc}(\rho(\mathbf{r})) = E_{xc}[\rho_{\mathbf{r}}] \tag{5}
$$
By now we have proved that your statement can only apply to LDA functionals, therefore definitely not to the exact density functional. However, most LDA functionals do not satisfy your statement either. To see this, consider two densities $\rho_A$ and $\rho_B$ such that
$$
\rho_A(\mathbf{r})=N\rho_B(\mathbf{r}) \quad \forall \mathbf{r} \tag{6}
$$
From Eq. (1) we have
$$
E_{xc}[\rho_A]=NE_{xc}[\rho_B] \tag{7}
$$
or
$$
\epsilon_{xc}(\rho_A(\mathbf{r}))=N\epsilon_{xc}(\rho_B(\mathbf{r})) \quad \forall \mathbf{r} \tag{8}
$$
This means
$$
\epsilon_{xc}(Nx)=N\epsilon_{xc}(x) \quad \forall x \tag{9}
$$
Obviously, only a linear functional of the type
$$
\epsilon_{xc}(x)=kx \tag{10}
$$
satisfies Eq. (9). Since the number of electrons of a system is the integral of the density over $\mathbb{R}^3$, we have that $E_{xc}$ is proportional to the number of electrons. This is such a bad approximation that no functional, not even the worst approximate ones, has this property. QED.