AC-to-DC Conversion: Which topology is better?
I am guessing the latter could be less efficient, however has fewer
components and no ICs, it will be cheaper and would require less area
on a PCB.

Yes it is far less efficient and yes it has fewer parts, but it may not be cheaper and probably won't take up less space. The only real advantages are that it uses generic parts that may be easier to source, and the design is less critical.
C1 must be rated for AC mains operation, so it will be much larger than an MLCC or electroytic type. According my calculations around 2.2 μF is required to get 50 mA at 5 V. At best (full load) the circuit could be around 60% efficient, but with no output it will consume almost double the power (since it has to absorb the current that would otherwise go to the load). This is very wasteful if the device spends most of its time in standby, as well as creating extra heat that may be hard to get rid of in a small enclosure.
The MP158 design needs 2 more components, but they are small compared to the size of the mains rated capacitor required for the second circuit. If you use a bridge rectifier then C1 can be even smaller. This circuit is much more efficient and draws minimal power on standby, so it should run a lot cooler.
Support components for the MP158 must be carefully chosen for proper operation, and the PCB layout is critical. However if you stick close to the recommended values and layout in the datasheet you should be OK. Obtaining supplies might be an issue (Mouser say it is 'obsolete'). You might consider using a later equivalent such as the MP173A.
To get some idea of how much space each design might take up, here are two similar examples taken from Texas Instruments design guide TIDA-010060:-

Both designs include the same post regulator, input protection devices and screw terminals, but the switching design takes up significantly less board area and less volume.