How to conclude the necessity for unbalanced to balanced signal conversion for this scenario?
The following applies to signal transmission where the operating frequencies of the signal have a wavelength that is much, much greater than the cable length.

A "balanced" output means that an output presents an impedance from each terminal to "earth" that is equal. This means that any EMI hitting the interconnecting cable spreads interference equally to both conductors. It then follows that a differential amplifier can eradicate this "common mode" noise interference by subtraction. But, of course, the input differential amplifier must have a balanced input impedance too; however, given that input impedances are usually very much higher than output impedances, this is less of a problem.

A "differential" output brings to the party reduced emitted interference from the cable providing that the system is balanced (impedance balanced). Because the signal amplitude is also double what an equivalent single-ended output has, there is usually a 2:1 improvement in signal-to-noise ratio.

So, in your final diagram, ideally you would have a 100 ohm resistor duplicated in the negative lead feeding the differential input. That will impedance-balance the cable and it means that any external EMI hitting the cable is spread equally on both conductors.

See this Q&amp;A for further details. This Q&amp;A is also relevant.