= Solution
Use $\hbar=1$ throughout. Let $N_a=a^\dagger a$, $N_b=b^\dagger b$ and $N=N_a+N_b$. Operators belonging to different <boson> species commute. The proposed <Schwinger boson representation> gives
$$
\begin{aligned}
S^+S^-&=a^\dagger b b^\dagger a=N_a(N_b+1),\\
S^-S^+&=b^\dagger a a^\dagger b=N_b(N_a+1).
\end{aligned}
$$
Consequently the first of the <spin commutation relations> is
$$
\boxed{[S^+,S^-]=N_a-N_b=2S^z.}
$$
The <number operator> <commutators> $[N_a,a^\dagger]=a^\dagger$, $[N_a,a]=-a$ and their $b$ counterparts give
$$
\begin{aligned}
[S^z,S^+]&=\tfrac12([N_a,a^\dagger]b-a^\dagger[N_b,b])=a^\dagger b=S^+,\\
[S^z,S^-]&=-b^\dagger a=-S^-.
\end{aligned}
$$
Thus all the <spin commutation relations> hold. Also $[N,S^\pm]=[N,S^z]=0$: each fixed-$N$ sector is invariant. To identify its <spin>, calculate the <Casimir operator>:
$$
\begin{aligned}
S^2&=(S^z)^2+\tfrac12(S^+S^-+S^-S^+)\\
&=\tfrac14(N_a-N_b)^2+N_aN_b+\tfrac12(N_a+N_b)\\
&=\frac N2\left(\frac N2+1\right).
\end{aligned}
$$
The <irreducible spin representation> with prescribed $s$ therefore requires $N=2s$, and its normalized <Fock basis> correspondence is
$$
\boxed{|s,m\rangle\ \longleftrightarrow\ |s+m,s-m\rangle
=\frac{(a^\dagger)^{s+m}(b^\dagger)^{s-m}}
{\sqrt{(s+m)!(s-m)!}}|0,0\rangle.}
$$
Conversely, $s=(n_a+n_b)/2$ and $m=(n_a-n_b)/2$. The nonnegative integer <occupation numbers> give precisely $m=-s,-s+1,\ldots,s$, and there are $2s+1$ such states. As a check, the <spin raising operator> acts as
$$
a^\dagger b|s+m,s-m\rangle
=\sqrt{(s+m+1)(s-m)}\,|s+m+1,s-m-1\rangle,
$$
which is the usual <spin ladder operator> coefficient $\sqrt{s(s+1)-m(m+1)}$.
For a lattice of fixed <spins>, the <local spin constraint in the Schwinger boson representation> is essential:
$$
\boxed{(a_n^\dagger a_n+b_n^\dagger b_n)|\psi\rangle=2s|\psi\rangle
\quad\text{at every site }n.}
$$
The full two-species <bosonic Fock space> also contains other local <irreducible spin representations>; fixing only the total <boson> number does not remove them.
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