Is indefinite integration non-linear?

The issue here is that the indefinite integral operator, when applied to a function, does not give a function as its output. It gives an equivalence class of functions, equivalent up to constant translation.

We usually write $\int 2x ~dx = x^2 +C$, but what we really mean is that $\int 2x ~dx = [x^2]$, where $$[x^2]=\{x^2+C: C\in \mathbb{R}\}.$$

Now, the apparent problem disappears: $$\int \overline{ 0} ~dx = [\overline{0}]$$ and $$0\int \overline{1} ~dx = 0[x] = [0x]=[\overline{0}]$$

In the above, $\overline{0}$ denotes the zero function, while $0$ denotes the zero scalar. (similarly $\overline{1}$ denotes the identically-one function). This notation is just for clarity, to emphasize the difference between the two zero symbols.

$\int f(x)dx$ is the usual notation for the set of all antiderivatives of $f(x).$ Thus, if $F(x)$ is a particular derivative of $f(x)$ then

$$\int f(x)dx=\{F(x)+c:c\in\mathbb{R}\},$$

if the domain of $f$ is connected. Thus,

$$\int \alpha f(x)dx=\{\alpha F(x)+c:c\in\mathbb{R}\}$$ if $\alpha\ne 0$ and

$$\int \alpha f(x)dx=\{c:c\in\mathbb{R}\}$$ if $\alpha=0.$

You are trying to translate linearity of definite integral to indefinite integral. But a definite integral gives you a number and an indefinite integral gives you a set of functions. So, linearity must be understood in this sense, and not with particular antiderivatives.