What calorimetry approach is this? Question: Find the final tempertature of the system when

The method used in the solution is to:
1.Bring all the constituents (in this case boiling hot steam and sub-zero ice) to a preselected and arbitrary point and calculate the heat flows (in or out) needed to achieve this for each component. In this case, you could pick all liquid water at $0^{\circ <!--\; \circ \; -->}$C.
2.Add up all the individual heat flows to find the net heat flow needed to reach this assumed condition.
3.Finally, realize that there should be zero heat flow. So apply the necessary correction to the single combined mass of water, to find its final temperature and state.
The problem with the method you prefer is that you might need to try many different final states to find the correct answer.
For a very small amount of steam, you might assume that the ice warms up slightly to $T$, while condensing the steam, cooling it from ${100}^{\circ <!--\; \circ \; -->}$ to $0^{\circ <!--\; \circ \; -->}$, freezing all of it, and cooling it from $0^{\circ <!--\; \circ \; -->}$ to $T$ below zero.
For a larger amount of steam, you might assume that the ice all warms up to $0^{\circ <!--\; \circ \; -->}$ while condensing all the steam, cooling it to $0^{\circ <!--\; \circ \; -->}$, and then freezing $x$ grams of the condensed steam.
For a still larger amount of steam, you might assume that the ice all warms up to $0^{\circ <!--\; \circ \; -->}$ while condensing all the steam, cooling it to $0^{\circ <!--\; \circ \; -->}$, and then freezing $x$ grams of the condensed steam.
For a still larger amount of steam, you might assume that the ice all warms up to $0^{\circ <!--\; \circ \; -->}$ and $x$ grams of it melts, while condensing all the steam and cooling it to $0^{\circ <!--\; \circ \; -->}$
For a still larger amount of steam you might assume that all the ice warms up to freezing, melts, and warms to $T$, while all the steam condenses and cools to $T$
And for a really large amount of steam, you might assume that all the ice warms to freezing, melts, warms up to ${100}^{\circ <!--\; \circ \; -->}$, while $x$ grams of the steam condenses.
Each of these is possible, depending on the amount of ateam, and you have no way of knowing which one to try just by looking at the numbers. You need to try each one. Each produces a different linear equation in one unknown, so each will have a single mathematical solution. But only one of the solutions will be consistent with the assumptions that went into finding it.
You might find values of $x$ that are negative, or too big for the amount of ice or steam actually present. You could assume all liquid water at $T$, and find that $T$ is ${125}^{\circ <!--\; \circ \; -->}$. You will need to keep trying scenario after scenario, to find the one that makes sense.

There is a difference between the masses in the question and those in the given answer. I assume the masses in the question are correct and those in the given answer are wrong.
In this "method" you imagine that the $5g$ of steam is converted to ice at $-<!--\; -\; -->{20}^{\circ <!--\; \circ \; -->}$C, so that it is the same as the $31.5g$ of ice which is already at $-<!--\; -\; -->{20}^{\circ <!--\; \circ \; -->}$C. The thermal energy extracted is then used to heat $31.5+5=36.5g$ of ice.
This method works because thermodynamic energy changes are conservative like changes in gravitational potential energy. The difference in energy depends only on the initial and final states and not on the path taken. In your method you have two initial states and you assume a temperature for the final state. In this method you convert everything to the same initial state (because it is the same substance) then use the energy extracted to heat it up to the final state.
If you are not happy with this method stick with your usual method, which will give you the same answer.