Let's say I have an oracle that outputs 1 on a specific input (there can be multiple inputs x so that f(x)=1), else it outputs 0.
If there is no such input so that f(x)=1, the quantum state of the output qubit will always be |0>
Is there a way so, that if the output state is not |0>, one can manipulate the qubit state as such that it transforms to |1>?
Not sure if I got your question exactly, so I reformulate it first. Assume some algorithm produces a state $$|\psi\rangle=\alpha |0\rangle+\beta|1\rangle.$$ It seems to me you are asking if there is a procedure to say if $\beta=0$ or not, i.e. if $|\psi\rangle$ is orthogonal to $|1\rangle$ or not. In other words, you want to discriminate two quantum states $|\psi\rangle$ and $|\phi\rangle=|0\rangle$. If you only have a single copy of $|\psi\rangle$ the answer is no, you can not do this with certainty.
Even if you know that the algorithm either produces $|\phi\rangle=|0\rangle$ or $|\psi\rangle$ (with known $\alpha$ and $\beta$) the best you can do is to tell them apart with with probability $1-|\langle\psi|\phi\rangle|$ https://en.wikipedia.org/wiki/POVM#An_example:_unambiguous_quantum_state_discrimination. Note that when the states are orthogonal $\langle\psi|\phi\rangle=0$ they can be discriminated with certainty.
@NikitaNemkov has provided a thorough answer to your question. I'd like to point out a simple reason as to why you cannot do that: namely, it violates both the unitarity as well as linearity of quantum mechanics.
Unitarity
A unitary process is reversible, i.e., it doesn't map two different states to the same state (otherwise, you cannot tell where the output state came from -- thus, making it irreversible). Now, it is obvious why the kind of operation you describe would be irreversible, i.e., non-unitary: just as an example, it would map both $\frac{1}{\sqrt{2}}(\vert{0}\rangle \pm\vert{1}\rangle)$ to $\vert{1}\rangle$.
Linearity
Linearity implies that if a process takes $\vert\psi_1\rangle$ to $\vert\phi_1\rangle$ and $\vert \psi_2\rangle$ to $\vert \phi_2\rangle$ then it ought to take $\frac{1}{\sqrt{2}}(\vert \psi_1\rangle +\vert \psi_2\rangle)$ to $\frac{1}{\sqrt{2}}(\vert \phi_1\rangle +\vert \phi_2\rangle)$. Now, the kind of operation you describe ought to take $\vert 0\rangle$ to $\vert 0\rangle$ and it ought to take $\vert 1\rangle$ to $\vert 1\rangle$. Thus, by linearity, it should take $\frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1\rangle)$ to $\frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1\rangle)$. However, since you want any state that is not $\vert 0\rangle$ to be mapped to $\vert 1\rangle$, your operation would rather want to map $\frac{1}{\sqrt{2}}(\vert 0 \rangle + \vert 1\rangle)$ to $\vert 1\rangle$ -- thus, violating linearity.
If that were possible, you could use it to communicate faster than light: Alice and Bob each hold one half of a Bell pair, and Alice measures in the computational basis if she wants to send a 0, and in the Hadamard basis if she wants to send a 1, while Bob measures $|0\rangle$ or not-$|0\rangle$. This channel has a high error rate (half of the 0 bits are flipped) but it can be made arbitrarily reliable with standard (classical) error correction techniques.
