Ta có: \({f}'\left( x \right)=\frac{1-m}{{{\left( x+1 \right)}^{2}}}\).

Nếu \(m=1\) thì \(f\left( x \right)=\frac{x+1}{x+1}=1,\forall x\ne -1\). Khi đó \(\underset{\left[ 0;1 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|+\underset{\left[ 0;1 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=2\) (thỏa mãn).

Do đó \(m=1\) thỏa mãn bài toán.

Nếu \(m\ne 1\) thì hàm số \(f\left( x \right)\) đơn điệu trên \(\left[ 0;1 \right]\).

• TH1: \(\left( \frac{m+1}{2} \right).m\le 0\) thì \(\underset{\left[ 0;1 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=0,\underset{\left[ 0;1 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\max \left\{ \frac{\left| m+1 \right|}{2};\left| m \right| \right\}\).

Do đó: \(\underset{\left[ 0;1 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|+\underset{\left[ 0;1 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=2\)\(\Leftrightarrow 0+\frac{\left| \frac{m+1}{2}+m \right|+\left| \frac{m+1}{2}-m \right|}{2}=2\)

\(\Leftrightarrow \frac{\left| 3m+1 \right|+\left| m-1 \right|}{4}=2\) \(\Leftrightarrow \left[ \begin{array}{l}
m > 1:m = 2\left( l \right)\\
1 > m \ge  - \frac{1}{3}:m = 3\left( l \right)\\
m <  - \frac{1}{3}:m =  - 2\left( l \right)
\end{array} \right.\). (so với điều kiện TH1)

• TH2: \(\left( \frac{m+1}{2} \right).m>0\) thì \(\underset{\left[ 0;1 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=\min \left\{ \frac{\left| m+1 \right|}{2};\left| m \right| \right\},\underset{\left[ 0;1 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|=\max \left\{ \frac{\left| m+1 \right|}{2};\left| m \right| \right\}\)

Do đó \(\underset{\left[ 0;1 \right]}{\mathop{\max }}\,\left| f\left( x \right) \right|+\underset{\left[ 0;1 \right]}{\mathop{\min }}\,\left| f\left( x \right) \right|=2\)

\(\Leftrightarrow \frac{\left| \left| \frac{m+1}{2}+m \right|-\left| \frac{m+1}{2}-m \right| \right|}{2}+\frac{\left| \frac{m+1}{2}+m \right|+\left| \frac{m+1}{2}-m \right|}{2}=2\)

\(\Leftrightarrow \frac{\left| \left| 3m+1 \right|-\left| m-1 \right| \right|}{4}+\frac{\left| 3m+1 \right|+\left| m-1 \right|}{4}=2\) \(\Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m =  - \frac{5}{3}
\end{array} \right.\)

Vậy \(S=\left\{ 1;\frac{-5}{3} \right\}\).