Analog switch on an instrumentation amplifier input
The analog switch by itself will not introduce a DC offset into the signal.  That is, it does not add electrons from its Vdd pin to the signal channel.  Leakage currents are sub-nanoamp in Fig. 9.  Only you can say whether or not this error is significant.

You don't say what the signal bandwidth or amplitude is, and the FET switch characteristics can affect both of those.

A single channel ON resistance varies between 40 and 125 ohms in the tables, but is worse in the chart for your conditions.  This is in series with all of your termination options.  For example, two of these channels are in series with your 50 ohm differential terminator, significantly affecting the effective termination impedance seen by the signal.

Datasheet fig. 7 shows how the resistance varies instantaneously with signal amplitude.  If the signals are source-terminated (example: 50 ohm output resistance at the source), this creates a voltage divider that varies with signal amplitude.  The series leg is the 50 ohms at the source, and the shunt leg varies between 210 and 260 ohms (50 ohms + (2 x switch resistance)).

Based on nothing but what you've said so far, I'd look for a better switch.  ADI and others have parts with 5 ohm (ish) ON values.