This sequent is provable.

\begin{logicproof}{2}
    (s \lor \lnot u) \imp t & \prem \\
    s \lor \lnot s & $\LEM$ \\
    \begin{subproof}
        s & \assum \\
        s\lor \lnot u & $\ori{1} 3$ \\
        t & $\impe 4, 1$ \\
        (\lnot s \land u) \lor t & $\ori{2} 5$
    \end{subproof}
    \begin{subproof}
        \lnot s & \assum \\
        u \lor \lnot u & $\LEM$ \\
        \begin{subproof}
            u & \assum \\
            \lnot s \land u & $\andi 7,9$ \\
            (\lnot s\land u) \lor t & $\ori{1} 10$
        \end{subproof}
        \begin{subproof}
            \lnot u & \assum \\
            s \lor \lnot u & $\ori{2} 12$ \\
            t & $\impe 13 1$ \\
            (\lnot s\land u) \lor t & $\ori{2} 14$
        \end{subproof}
        (\lnot s\land u) \lor t & $\ore 8, 9-11, 12-15$
    \end{subproof}
    (\lnot s\land u) \lor t & $\ore 2, 3-6, 7-16$
\end{logicproof}