Are these two reactions equivalent?
Let's start with the question: How do we get from your answer to the official answer?  We can transform your answer into the official answer by first adding 3 ammonium ions to each side:

$\begin{aligned}\ce{H2O} + \ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} &amp; \rightarrow \ce{NH3} + 2\ce{S} + 3\ce{OH-} \\
+ 3 \ce{NH3}+3\ce{H^+} &amp; \rightarrow + 3 \ce{NH3}+3\ce{H^+}\\
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} &amp; \rightarrow 4\ce{NH3} +3\ce{H^+} + 2\ce{S} + 3\ce{OH-}
\end{aligned}$

Then noting that the $\ce{H+}$ and $\ce{OH-}$ will 'neutralize' one another on the product side, producing $\ce{3H2O}$, one of which will 'cancel' the water molecule on the reagent side:

$\begin{aligned}
\ce{H2O} + 4\ce{NH4+} + 2\ce{S^{2-}} + \ce{O2} &amp; \rightarrow 4\ce{NH3} + 2\ce{S} + 3\ce{H2O} \\
-\ce{H2O} &amp; \rightarrow - \ce{H2O}\\
4\ce{NH4+}+2\ce{S^2-}+\ce{O2} &amp; \rightarrow 4\ce{NH3}+2\ce{S}+2\ce{H2O}
\end{aligned}$

Producing what is the 'official' result.  Now we can ask: are the two answers equivalent?  Because $\ce{NH4+}$ is a weak acid in water, it will dissociate and we can write your answer as:

$\ce{H2O + NH3 + H+ +2S^2- + O2} \rightarrow \ce{NH3 + 2S + 3OH-}$

Your answer is a balanced redox reaction somewhere in between acidic and basic conditions, which is a rather unusual formulation.  I'm not going to go so far as to say, this can't happen, but it is unlikely, and certainly not how we typically describe balanced redox reactions.