= Absolute pressure in Boyle's law
With <gauge pressure> $p_g$ referenced to <atmospheric pressure> $p_{\rm atm}$, <absolute pressure> is $p=p_g+p_{\rm atm}$. Thus <Boyle's law> becomes
$$
(p_{g,1}+p_{{\rm atm},1})V_1
=(p_{g,2}+p_{{\rm atm},2})V_2.
$$
Even a constant <atmospheric pressure> cannot be dropped from these sums. For the calculated $100\,\mathrm{kPa}$ to $200\,\mathrm{kPa}$ example, take $p_{\rm atm}=100\,\mathrm{kPa}$. The initial and final <gauge pressures> are $0$ and $100\,\mathrm{kPa}$; their products with the respective <volume (thermodynamics)> are $0$ and $5000\,\mathrm{kPa\,cm^3}$, which are unequal. Zero <gauge pressure> means balance with the surroundings, not absence of <gas> or absence of wall <momentum> transfer.
For a frictionless piston of <surface area> $A$ with its outer face exposed to <atmospheric pressure>, the additional inward <force> required for <mechanical equilibrium> is $F=(p-p_{\rm atm})A=p_gA$. This is why a mechanical load can reveal a <pressure> difference even though the <equation of state> requires <absolute pressure>. With piston <friction>, the load alone no longer determines $p_gA$; a direct <pressure sensor> avoids that particular inference.
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