After having looked into the history of calculus and into original works by Leibniz and others, I have come to the conclusion that the schizophrenic way the "indefinite integral" is sometimes taught is a byproduct of the common schizophrenic mixture of Leibniz-time notation with Bourbaki-time foundations of calculus. I've found a satisfactory way to deal with the former, but not yet with the latter (I am working on it).
When trying to make sense of the traditional Leibniz notation in calculus, it is important to understand that even the notion of function in the 18th century was not what it is now. Now a function is something along the lines of "a set of pairs," "a functional relation," "a triple of a functional relation, a domain, and a co-domain," "a morphism in the category of sets." In the 18th century, to be a function was a property/predicate on variables: one variable could be a function of another, and also two variables could be functions of each other.
According to what I have understood, the notation of Leibniz, as used in the 18th century, heavily relied on the context. Expressions, like integrals, were not self-contained. In fact, even integration bounds were not written below and above "$\int$." Apparently, the first well-known text where the bounds of integration appeared at the integration symbol was Théorie analytique de la chaleur (1822) by Joseph Fourier -- see The history of notations of the Calculus (1923) by Florian Cajori.
I long believed that in the expression "$\int x^2dx$," the variable "$x$" is bound (is not free) because of "$dx$," but this is not at all the idea. One can have "$\int ydx$," "$\int xdy$," "$\int(xdy + ydx)$," and how this is to be integrates should be specified separately. For instance:
$$
\int_{x=0}^1 x^2dx =\frac{1}{3},\qquad
\int_{y=0}^1 x^2dx = 0,
$$
where in the second integral the variable "$x$" represents a fixed quantity (fixed in the context of this integral), while $y$ varies from $0$ to $1$.
(This is not how this would have been written in the 18th century, because bounds were not written by the integration symbol, but this is the idea.)
Now, I return to the question of the meaning of the "indefinite integral."
If we simply want to write identities like
$$
\int (f(x) + g(x))dx =\int f(x)dx +\int g(x)dx
$$
without worrying about the bounds, like this would have been done in the 18th century, it suffices to interpret the "indefinite integrals" that appear here as indeterminate definite integrals -- definite integrals with indeterminate bounds. The meaning of this identity then is that for all $a$ et $b$ such that both sides are defined, we have:
$$
\int_{x=a}^b (f(x) + g(x))dx =\int_{x=a}^b f(x)dx +\int_{x=a}^b g(x)dx.
$$
In modern terms, this amounts to an identity of functions of two variables, which take $a$ et $b$ and return the value obtained by integrations from $a$ to $b$. I believed having already seen such a definition of indefinite integral given by Paul Halmos, but I have just tried to find it, and I have no idea where I could have seen it. Perhaps I deduced it myself from the definition of indefinite Lebesgue integral given by Halmos (see below).
Here is the definition of an indefinite integral that I plan to use in my teaching from now on:
$\int_x f(x)dx$, written also as "$\int f(x)dx$" when it is clear from the context that the integration is with respect to $x$, is the function that takes a pair of reals $(a, b)$ and returns the value $\int_{x=a}^b f(x)dx$. We adopt the convention to not write expressions like "$(\int_x f(x)dx)(0, 1)$," but to write instead "$\int_{x=0}^1 f(x)dx$."
If teaching time permits, I may mention that historically there was another notion of "indefinite integral," which I may call, for example, indefinite integral in the sense of Cauchy (see my another partial answer), but that this historical notion is virtually informalizable in the modern mathematics.
In addition to the proposed definition of an indefinite integral, I plan to use the following definition of an indefinite difference:
$[f(x)]_x$, written also as "$[f(x)]$" when there should be no ambiguity, is the function that takes a pair of reals $(a, b)$ and returns the value $[f(x)]_{x=a}^b = f(b) - f(a)$. We adopt the convention to not write expressions like "$([f(x)]_x)(0, 1)$," but to write instead "$[f(x)]_{x=0}^1$."
We have the following obvious property:
$$
[f(x)] = [g(x)]\quad\Leftrightarrow\quad
[f(x) - g(x)] = [0]\quad\Leftrightarrow\quad
f - g\ \ \text{is constant}.
$$
The second fundamental theorem of calculus can now be written as:
$$
\int f'(x)dx = [f(x)].
$$
This interpretation is compatible with and can be applied to "non-oriented" integrals $\int f(x)|dx|$, Stieltjes integrals $\int f(x)dg(x)$, "non-oriented" Stieltjes integrals $\int f(x)|dg(x)|$, etc.
This interpretation also agrees with my recent idea to view measure-theoretic integration as a binary operation that takes a signed measure and a measurable function and returns a signed measure. Just now I've re-discovered that this is exactly the definition of indefinite Lebesgue integral given by Paul Halmos in his Measure theory. Thus, the indefinite Lebesgue integral $\int fd\mu$ is the signed measure that takes a measurable set $X$ and returns the number $\int_X fd\mu$.
I am adding some examples of calculations:
$$
\int_t te^t dt =\int_t tde^t
= [te^t]_t -\int_t e^tdt = [te^t]_t -[e^t]_t = [te^t - e^t]_t.
$$
\begin{multline*}
\int_x\frac{dx}{1 - x^2}
=\int_x\frac{1}{2}\left(\frac{dx}{1 - x} +\frac{dx}{1 + x}\right)
=\frac{1}{2}\left(\int_x\frac{dx}{x + 1} -\int_x\frac{dx}{x - 1} \right)\\
=\frac{1}{2}([\ln |x + 1|]_x -[\ln |x - 1|]_x)
=\frac{1}{2}[\ln |x + 1| -\ln |x - 1|]_x
=\left[\frac{1}{2}\ln\left|\frac{x + 1}{x - 1}\right|\right]_x.
\end{multline*}
In the second example, it is by default understood that the obtained identity is equivalent to
$$
\int_{x=a}^b\frac{dx}{1 - x^2}
=\left[\frac{1}{2}\ln\left|\frac{x + 1}{x - 1}\right|\right]_{x=a}^b
$$
for all $a,b\in\mathbf{R}$ such that the integral exists, which excludes the possibility that $a$ and $b$ be separated by $1$ or $-1$.
