2:1 coupler is not a 2:2 coupler with one input chopped off. Not quite, at least. If you send two optical pulses, in phase, into the two inputs of a 2:1 coupler, they will be combined into a single mode on the other side. If, however the two signals are out of phase, the two signals will not be combined.

The best way to think about it is this. Lets say all outputs/inputs of your 2:1 coupler are identical single-mode waveguides, and let us ignore polarization degree of freedom for now.

Lets say the two waveguides on one side of your waveguide are A and B, and a single waveguide on the other side is S.

The field is the S-waveguide is then: $E_s=(E_A+E_B)/\sqrt{2}$, i.e. S takes in the symmetric part of your AB-input, the other anti-symmetric part $E_A - E_B$ is the bit that is lost. If your AB-input is purely summetric, i.e. $E_A$ and $E_B$ are equal in phase and amplitude, then no energy will be lost.

This is all classical optics however. You will need to run more calculations to see what happens, given boson statistics, if $E_A$ and $E_B$ are single-photon states.

So in answer to your question. I think a 2:1 coupler could still do the job, but you need to tell more about your specific setup. Can you get $E_A$ and $E_B$ to be in phase and of the same amplitude?

If you send two optical pulses, in phase, into the two inputs of a 2:1 coupler, they will be combined into a single mode on the other side. If, however the two signals are out of phase, the two signals will not be combined.

The best way to think about it is this. Lets say all outputs/inputs of your 2:1 coupler are identical single-mode waveguides, and let us ignore polarization degree of freedom for now.

Lets say the two waveguides on one side of your waveguide are A and B, and a single waveguide on the other side is S.

The field is the S-waveguide is then: $E_s=(E_A+E_B)/\sqrt{2}$, i.e. S takes in the symmetric part of your AB-input, the other anti-symmetric part $E_A - E_B$ is the bit that is lost. If your AB-input is purely summetric, i.e. $E_A$ and $E_B$ are equal in phase and amplitude, then no energy will be lost.

This is all classical optics however. You will need to run more calculations to see what happens, given boson statistics, if $E_A$ and $E_B$ are single-photon states.

So in answer to your question. I think a 2:1 coupler could still do the job, but you need to tell more about your specific setup. Can you get $E_A$ and $E_B$ to be in phase and of the same amplitude?

Following the comment.

1. I was to rash to say it is not a beam splitter with one output ignored. Indeed, the output of the 2:1 coupler is one of the standard outputs of a beam splitter. I have corrected it

2. The $E_A - E_B$ part will be coupled out of the single-mode '1' waveguide (in the 2:1 coupler), and will therefore be lost to environment, be it housing of the coupler or free-space. I remember seeing papers, published in 1970s or 1980s actually showing detailed derivaions of this, but I cannot find them.

3. In the most general case, I don't think you will be able to merge the two optical modes, at least linearly. Are your signals of definite polarization? If they are, you could mix them via a polarization-sensitive beam-splitter. Of course, if two signals are out of phase, but phase relation-ship is stable, an optical delay line may fix the problem, though I doubt it is that simple. Another option would be to see if non-linear approaches could be harnessed, but then you need a crystal, a pump, etc.

Another option is post-selection. As I understand, in quantum optics it is common to have schemes that do not always work, but work a certain known share of the time, and this is fine as long as one can decide on detection whether the scheme has worked that time. If you do indeed have to single-photon states incident on two different arms of a beam splitted, AFAIK, due to HOM effect those two photons will exit as a two-photon state out of one of the arms of the beam-splitter. So you can have a scheme where there is a conventional 2:2 coupler with inputs A, B and outputs S, X. You send two single-photon states into A and B and make sure their envelopes overlap as well as possible. Then you arrange your optical experiment to be fed by output S, and put a detector on output X.The procedure is then to disregard the optical experiment if detector X clicks, and if it does not click, then you know that, that time, the two photons on the inputs have existed as a two-photon state from output S.