The answer to the question is as follows:
Theorem 1. For every $C > 0$ there is, up to scalar multiples, only one function $0 \le f \in L^1 := L^1((0,\infty))$ such that $Tf = f$ (where $T$ is the $C$-dependent operator from the question).
For the proof we need a few preparations.
First we quote the following result (which is actually true for all Banach lattices with order continuous norm, but we restrict ourselves to $L^p$-spaces here); in what follows, all inequalites are to be understood almost everywhere.
Theorem 2. Let $(\Omega,\mu)$ be an arbitrary measure space and let $p \in [1,\infty)$. Let $0 \le u \in L^p := L^p(\Omega,\mu)$ and suppose that $S$ is a bounded linear operator on $L^p$ such that for each $f \in L^p$ there exists a number $c \ge 0$ for which we have $\lvert Sf\rvert \le c u$. Then:
Theorem 2 is a consequence of standard results about Dunford-Pettis operators on Banach lattices which can for instance be found in [P. Meyer-Nieberg: Banach Lattices (1991)]; however, I don't know an explicit reference for Theorem 2 itself in a classical monograph. An explicit reference for Theorem 2 in a recent article (including a proof and detailed references) is [D. Daners, J. Glück: The Role of Domination and Smoothing Conditions in the Theory of Eventually Positive Semigroups, Theorem 2.3] (where the result is formulated in the more general setting of Banach lattices).
As a consequence of the general Theorem 2 we obtain the following result for the operator $T$ in our concrete situation:
Proposition 3. The operator $T^2$ is compact from $L^1$ to $L^1$. In particular, the spectrum of $T$ is at most countable, and all spectral valued of $T$, except for possibly the number $0$, are eigenvalues of finite algebraic multiplicity and poles of the resolvent of $T$.
Proof. For every $f \in L^1$ we have $\lvert Tf\vert \le T\lvert f \vert \le C \lVert f\rVert_{L^1} u$, where $u \in L^1$ is the function given by $u(x) = 1/(1+x)^2$. Thus, the assertion follows from Theorem 2.
Remark. Mateusz Kwaśnicki noted in his answer that $T$ is compact (even a Hilbert-Schmidt operator) from $L^2$ to $L^2$, but Proposition 3 above deals with the operator on $L^1$.
Now we are going to use infinite dimensional Perron-Frobenius theory. We first recall a few notions for readers not familiar with this theory:
Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, let $p \in [1,\infty)$; we set $L^p := L^p(\Omega,\mu)$.
We call a function $f \in L^p$ positive and denote this by $f \ge 0$ if $f(\omega) \ge 0$ for almost all $\omega \in \Omega$.
Let $S$ be a bounded linear operator on $L^p$.
The operator $S$ is called positive if $Sf \ge 0$ for all $f \ge 0$.
The operator $S$ is called reducible if there exists a measurable set $M \subseteq \Omega$ such that both $M$ and $\Omega \setminus M$ have non-zero measure and such that $Sf$ vanishes almost everywhere on $M$ whenever $f$ vanishes almost everywhere on $M$.
The operator $S$ is called irreducible if it is not reducible.
Remarks. (a) All the above notions can also be defined in the more general context of Banach lattices, and the subsequent theorem is also true on Banach lattices; however, I think this post might be more accessible if we restrict our attention to $L^p$-spaces).
(b) In particular, one can also define reducible and irreducible operators on $L^p$ if $(\Omega,\mu)$ is not $\sigma$-finite (since $L^p$ is still a Banach lattice in this case); however, the above way to define those notions is not appropriate in this setting; instead one has to use more abstract notions from Banach lattice theory to get a reasonable theory of irreducible operators on $L^p$-spaces over non-$\sigma$-finite measure spaces.
Now we quote the following result which belongs to what is usually called infinite-dimensional Perron--Frobenis theory:
Theorem 4. Let $(\Omega,\mu)$ be a $\sigma$-finite measure space and let $S$ be a positive and irreducible bounded linear operator on $L^p := L^p(\Omega,\mu)$ of spectral radius $r(S) = 1$. If some power of $S$ is compact, then the following assertions hold:
(a) The fixed space $\ker(1-S)$ is one-dimensional.
(b) The number $1$ is the only eigenvalue of $S$ with a positive eigenvector.
This result can, for instance, be found in [H. H. Schaefer: Banach Lattices and Positive Operators (1974), Theorem V.5.2 and part (ii) of this theorem's corollary].
Now we can prove Theorem 1:
Proof of Theorem 1. First we note that the operator $T$ is irreducible since we have $Tf(x) > 0$ for almost all $x \in [0,\infty)$ whenever $0 \not= f \ge 0$.
Let $r(T) \in [0,\infty)$ denote the spectral radius of $T$. We distinguish between three case:
First case: $r(T) < 1$. In this case we obviously have $\ker(1 - T) = \{0\}$.
Second case: $r(T) = 1$. In this case the space $\ker(1-T)$ is one-dimensional according to Theorem 4(a).
Third case: $r(T) > 1$. By applying Theorem 4(b) to the operator $S := T/r(T)$ we can see that $r(T)$ is the only eigenvalue of $T$ with a positive eigenvector. Hence, $\ker(1-T)$ does not contain any non-zero functions $f$ which fulfil $f \ge 0$. This completes the proof.
Remark. Actually, I'm not sure whether the spectral radius of $T$ is really non-trivial or whether we have $r(T) = 0$ (which is true for some $C$ if and only if it is true for all $C$).