Here is another take at that problem.

First let's define some quantities that will be useful later:

\[ \begin{align} A &= \frac {V_2} {V_1} \qquad\qquad B = \frac {R_L} {R_S} \qquad\qquad r_1 = \frac {R_1} {R_L} \qquad\qquad r_2 = \frac {R_2} {R_L} \qquad\qquad r_3 = \frac {R_3} {R_L} \tag{1} \end{align} \]

Since the attenuator port 1 has an input impedance that matches the source, then the source voltage is split in two at the input. Moreover, we also apply the definition of attenuation $A$.

\[ \begin{align} V_1 &= \frac{V_S}{2} \qquad\qquad V_2 = A V_1 = A \frac{V_S}{2} \tag{2} \end{align} \]

We can therefore compute the currents into the ports as:

\[ \begin{align} I_1 &= \frac{V_S}{2R_S} \tag{3} \\[2 em] I_2 &= -\frac{V_2}{R_L} = -\frac{A}{R_L} \frac{V_S}{2} \tag{4} \end{align} \]

Now let's focus on the internal node, the junction between $R_1,R_2, R_3$ (let's call it node 3). You can use KCL to compute the current in $R_3$ and then use Ohm's law to obtain $V_3$:

\[ \begin{align} I_3 &= I_1 + I_2 = \frac{V_S}{2R_S} - \frac{A}{R_L} \frac{V_S}{2} = \biggl(\frac{1}{R_S} - \frac{A}{R_L} \biggr) \frac{V_S}{2} \tag{5} \\[2 em] V_3 &= R_3 I_3 = R_3 \biggl(\frac{1}{R_S} - \frac{A}{R_L} \biggr) \frac{V_S}{2} = \frac{R_3}{R_L} \biggl(\frac{R_L}{R_S} - A \biggr) \frac{V_S}{2} = r_3 \biggl(B - A \biggr) \frac{V_S}{2} \tag{6} \end{align} \]

Now we can compute $V_3$ also using KVL applied to the input and output loops, obtaining two more independent equations for $V_3$:

\[ \begin{align} V_3 &= V_1 - R_1 I_1 = \frac{V_S}{2} - R_1 \frac{V_S}{2 R_S} = \biggl(1 - \frac{R_1}{R_S} \biggr) \frac{V_S}{2} = \biggl(1 - \frac{R_L}{R_S} \cdot \frac{R_1}{R_L} \biggr) \frac{V_S}{2} = \biggl(1 - B \cdot r_1 \biggr) \frac{V_S}{2} \tag{7} \\[2 em] V_3 &= V_2 - R_2 I_2 = \frac{A V_S}{2} - R_2 \biggl(-\frac{A}{R_L} \frac{V_S}{2} \biggr) = A \biggl(1 + \frac{R_2}{R_L} \biggr) \frac{V_S}{2} = A \biggl(1 + r_2 \biggr) \frac{V_S}{2} \tag{8} \end{align} \]

Now comparing equation 7 with equations 8 and 9, respectively, we obtain two independent equations in which the only unknown are $r_1,r_2,r_3$:

\[ \begin{align} \left\{ \begin{aligned} r_3 \biggl(B - A \biggr) \frac{V_S}{2} = \biggl(1 - B \cdot r_1 \biggr) \frac{V_S}{2} \\[2 em] r_3 \biggl(B - A \biggr) \frac{V_S}{2} = A \biggl(1 + r_2 \biggr) \frac{V_S}{2} \end{aligned} \right. \qquad\qquad &\Leftrightarrow \qquad\qquad \left\{ \begin{aligned} r_3 \biggl(B - A \biggr) = \biggl(1 - B \cdot r_1 \biggr) \\[2 em] r_3 \biggl(B - A \biggr) = A \biggl(1 + r_2 \biggr) \end{aligned} \right. \qquad\qquad \Leftrightarrow \qquad\qquad \nonumber\\[2 em] \qquad\qquad &\Leftrightarrow \qquad\qquad \left\{ \begin{aligned} r_1 = \frac{1 - r_3 \biggl(B - A \biggr)}{B} \\[2 em] r_2 = \frac{r_3 \biggl(B - A \biggr) - A}{A} \end{aligned} \right. \tag{9} \end{align} \]

Now let's focus on port 2. Looking into it we see an impedance $R_L$ which can be calculated using the usual methods of Thevenin's theorem:

\[ \begin{gather} R_L = R_2 + R_3 \parallel (R_1 + R_S) = R_2 + \frac{1}{ \frac{1}{R_3} + \frac{1}{R_1 + R_S} } \qquad \Leftrightarrow \tag{10} \\[2 em] \quad \Leftrightarrow \quad R_L - R_2 = \frac{1}{ \frac{1}{R_3} + \frac{1}{R_1 + R_S} } \quad \Leftrightarrow \quad \frac{1}{ R_L - R_2 } = \frac{1}{R_3} + \frac{1}{R_1 + R_S} \quad \Leftrightarrow \quad \frac{1}{ 1 - \frac{R_2}{R_L} } = \frac{R_L}{R_3} + \frac{1}{\frac{R_1}{R_L} + \frac{R_S}{R_L}} \quad \Leftrightarrow \nonumber \\[2 em] \qquad \Leftrightarrow \qquad \frac{1}{ 1 - r_2 } = \frac{1}{r_3} + \frac{1}{r_1 + \frac{1}{B}} \tag{11} \end{gather} \]

Now let's rewrite equations (9) like this:

\[ \begin{gather} \left\{ \begin{aligned} r_1 = \frac{1 - r_3 \biggl(B - A \biggr)}{B} \\[2 em] r_2 = \frac{r_3 \biggl(B - A \biggr) - A}{A} \end{aligned} \right. \qquad \Leftrightarrow \qquad \left\{ \begin{aligned} r_1 + \frac{1}{B} = \frac{2 + r_3 \biggl(A - B \biggr)}{B} \\[2 em] 1 - r_2 = \frac{2A - r_3 \biggl(B - A \biggr)}{A} \end{aligned} \right. \qquad \Leftrightarrow \qquad \left\{ \begin{aligned} \frac{1}{ r_1 + \frac{1}{B} } = \frac{B}{2 + r_3 \biggl(A - B \biggr)} \\[2 em] \frac{1}{ 1 - r_2 } = \frac{A}{2A + r_3 \biggl(A - B \biggr)} \end{aligned} \right. \tag{12} \end{gather} \]

substituting (12) into (11) we get:

\[ \begin{gather} \frac{1}{ 1 - r_2 } = \frac{1}{r_3} + \frac{1}{r_1 + \frac{1}{B}} \qquad \Leftrightarrow \qquad \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{1}{r_3} + \frac{B}{2 + r_3 \biggl(A - B \biggr)} \tag{13} \end{gather} \]

Now we can solve (13) for $r_3$.

\[ \begin{gather} \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{1}{r_3} + \frac{B}{2 + r_3 \biggl(A - B \biggr)} \qquad \Leftrightarrow \qquad \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{2 + r_3 \biggl(A - B \biggr) + B r_3}{r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]} \nonumber \\[2 em] \Leftrightarrow \qquad \frac{A}{2A + r_3 \biggl(A - B \biggr)} = \frac{2 + A r_3 }{r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr]} \qquad \Leftrightarrow \nonumber \\[2 em] \qquad A r_3 \biggl[ 2 + r_3 \biggl(A - B \biggr) \biggr] = (2 + A r_3 ) \biggl[ 2A + r_3 \biggl(A - B \biggr) \biggr] \qquad \Leftrightarrow \nonumber \\[2 em] \qquad 2A r_3 + A r_3^2 \biggl(A - B \biggr) = 4A + 2r_3 \biggl(A - B \biggr) + 2A^2 r_3 + A r_3^2 \biggl(A - B \biggr) \nonumber \\[2 em] \qquad 0 = 4A - 2Br_3 + 2A^2 r_3 = 4A + 2 r_3 (A^2 - B ) \qquad \Leftrightarrow \qquad r_3 (B - A^2 ) = 2A \nonumber \\[2 em] r_3 = \frac {2A}{ B - A^2 } \tag{14} \end{gather} \]

And then substitute $r_3$ back into equations (12) to get $r_1,r_2$:

\[ \begin{gather} \left\{ \begin{aligned} r_1 &= \frac{1 - r_3 \biggl(B - A \biggr)}{B} = \frac{1 - \frac {2A}{ B - A^2 } \biggl(B - A \biggr)}{B} = \frac{B - A^2 - 2A \biggl(B - A \biggr)}{B (B - A^2)} \\[2 em] r_2 &= \frac{r_3 \biggl(B - A \biggr) - A}{A} = \frac{ \frac {2A}{ B - A^2 } \biggl(B - A \biggr) - A}{A} = \frac{ 2A \biggl(B - A \biggr) - A (B - A^2)}{A (B - A^2)} \end{aligned} \right. \qquad \Leftrightarrow \nonumber \\[2 em] \Leftrightarrow \qquad \left\{ \begin{aligned} r_1 &= \frac{B - A^2 - 2A \biggl(B - A \biggr)}{B (B - A^2)} = \frac{B + A^2 - 2AB}{B (B - A^2)} = \frac{B ( 1 + A^2/B - 2A)}{B (B - A^2)} \\[2 em] r_2 &= \frac{ 2A \biggl(B - A \biggr) - A (B - A^2)}{A (B - A^2)} = \frac{ AB + A^3 - 2A^2}{A (B - A^2)} = \frac{ A (B + A^2 - 2A)}{A (B - A^2)} \end{aligned} \right. \qquad \Leftrightarrow \nonumber \\[2 em] \Leftrightarrow \qquad \left\{ \begin{aligned} r_1 &= \frac{B ( 1 + A^2/B - 2A)}{B (B - A^2)} = \frac{ 1 + A^2/B - 2A}{B - A^2} \\[2 em] r_2 &= \frac{ A (B + A^2 - 2A)}{A (B - A^2)} = \frac{ B + A^2 - 2A}{B - A^2} \end{aligned} \right. \tag{15} \end{gather} \]

Therefore we end up with our solution for $r_1,r_2,r_3$, from which you can get $R_1,R_2,R_3$:

\[ \begin{gather} \left\{ \begin{aligned} r_1 &= \frac{ 1 + A^2/B - 2A}{B - A^2} \\[2 em] r_2 &= \frac{ B + A^2 - 2A}{B - A^2} \\[2 em] r_3 &= \frac {2A}{ B - A^2 } \end{aligned} \right. \qquad \Leftrightarrow \qquad \left\{ \begin{aligned} R_1 &= R_L \cdot \frac{ 1 + A^2/B - 2A}{B - A^2} \\[2 em] R_2 &= R_L \cdot \frac{ B + A^2 - 2A}{B - A^2} \\[2 em] R_3 &= R_L \cdot \frac {2A}{ B - A^2 } \end{aligned} \right. \tag{16} \end{gather} \]

which matches your solutions.