Adi emailed me this answer, so I will post it.
Let us decompose $X=W\oplus E$ for some finite-dimensional subspace $E$ of $X$. Denote by $P_W\in\mathcal{B}(X)$ the bounded linear projection onto $W$ along $E$, and by $E^\perp\subseteq X^*$ the annihilator of $E$. Define the bounded linear map $j:E^\perp\to W^*$ by the rule \begin{equation}jx^*=x^*|_W\;\;\;\text{ for all }x^*\in E^\perp,\end{equation} which admits a bounded linear inverse $j^{-1}:W^*\to E^\perp$ given by the rule \begin{equation}j^{-1}w^*=w^*\circ P_W\;\;\;\text{ for all }w^*\in W^*.\end{equation}
Set $R:=j^{-1}\circ S^{*}\circ j:E^{\perp}\to E^{\perp}$, and let $H$ be an AIHS for $S^{*}$ with error $M$. Put $G:=j^{-1}H$ and $N:=j^{-1}M$, which are closed subspaces of $E^{\perp}$ and hence also of $X^{*}$. Then for any $x^{*}\in E^{\perp}$ and any $x\in X$ we now have \begin{eqnarray*} ((P_W^{*}T^{*})x^{*})(x) & = & x^{*}(TP_Wx) \hskip 4cm (\mbox{note } P_Wx\in W)\\ &=& x^{*}(SP_Wx)=(jx^{*})(SP_Wx) \hskip 1cm (\mbox {note } (jx^{*})\in W^{*} )\\ &=& (S^{*}(jx^{*}))(P_Wx) = ((j^{-1}\circ S^{*}\circ j)(x^{*}))(x) \\ &=& (Rx^{*})(x) \end{eqnarray*} so that $P_W^{*}T^{*}\equiv R$ on $E^\perp$. Since also $G\subseteq E^{\perp}$, this gives us \begin{multline*}(P_W^{*}T^{*})G=RG=(j^{-1}\circ S^{*}\circ j)(j^{-1}H)\\=j^{-1}(S^{*}H)\subseteq j^{-1}(H+M)=G+N,\end{multline*} and hence \begin{equation}\tag{1}T^{*}G=(P_W^{*}T^{*})G+((1-P_W^{*})T^{*})G\subseteq G+N+((1-P_W^{*})T^{*})G.\end{equation} Furthermore, since $P_W$ is a projection we have $(P_W^*T^*)(h\circ P_W)=T^*(h\circ P_W)$ and hence \begin{multline*}((1-P_W^*)T^*)(j^{-1}h)=((1-P_W^*)T^*)(h\circ P_W)\\=T^*(h\circ P_W)-(P_W^*T^*)(h\circ P_W)=0.\end{multline*} for every $h\in H$. As any element of $G$ has the form $j^{-1}h$ for some $h\in H$ this means $((1-P_W^*)T^*)G=\{0\}$. Together with $(1)$, this means \begin{equation}T^*G\subseteq G+N.\end{equation} Thus, the space $G$ is an AIHS under $T^*$ with error of dimension $\leq\text{dim}(N)=\text{dim}(M)$.