This is a bizarre question, but here goes...
If all of the water in the oceans were boiled into steam by the newly forming molten earth, could the atmosphere retain the steam? In other words, how much "space" does the water of the oceans occupy? If converted into steam, would the "space" of the atmosphere be great enough to contain the expanded water vapors?
Using mostly geometry for estimation:
Earth radius is about $r = 6400 \mbox{ km}$. Water covers about $70\%$ of the earth surface, estimation: $A_w = 4 \pi r^2 = 360,302,978 \mbox{ km}^2$
Wikipedia lists $360,570,000 \mbox{ km}^2$.
Average ocean depth is $d = 3.8 \mbox{ km}$. So a crude estimate would be a water volume of about $V_w = A_w d = 1,369,151,317 \mbox{km}^3$.
I found an estimation of the water volume as $V_w = 1,386,000,000 \mbox{km}^3$.
A mol of water weighs about $18\mbox{ g}$ this is a $18 \mbox{ ml}$ volume. As a gas it would take about 22 litres (depending on temperature and pressure). That is roughly an increase in volume about $1200$ times.
How much should we extend the earth radius for this? $$ 1200 V_w = \frac{4}{3}\pi (r + dr)^3 - \frac{4}{3}\pi dr^3 = \frac{4}{3}\pi (r^3 + 3r^2 \, dr + O(dr^2)) \Rightarrow \\ dr \approx \frac{\frac{3}{4} \frac{1200 V_w}{\pi} - r^3}{3r^2} = 1059 \mbox{ km} $$ Wikipedia lists that about $75\%$ of the atmosphere is contained within $11 \mbox{ km}$. The space border is about $100 \mbox{ km}$ (Kármán line).
So this would need a much larger volume. In reality the temperature is low (shrinks the gas volume) as is the pressure (expands the gas volume). However I saw only compression factors in the range of 1 to 10 for low temperatures.
So my crude guess is that the needed volume to contain the atmosphere plus water as some form of steam would be vastly greater than what our present atmosphere contains.