Comment to your attempt: instead of reasoning based on the initial positions and conditions of the dog and cat, I advise to sketch a generic intermediate position of both. I tried this approach in the sketch below.
I have not checked your second computation whose result differs from mine. I think you should explain it with some text and a sketch.
I arrived at the same result of Egor Skriptunoff's answer but have not tried to compute and analyze the invariant expression stated in his answer.
I assume that: (a) the dog starts at the point $S=(L,0)$ and the cat at the origin $O=(0,0)$; (b) the cat moves in the positive direction along the $y$-axis, and the dog describes a curve of pursuit (WolframMathWorld link) $C$ in the $xy$-plane. I call $y=f(x)$ the equation of $C$.

At time $t$ the tangent line to $C$ at the point $P(x,y)$ passes through the point $Q=(0,ut)$, which means that the derivative $y^{\prime }=f^{\prime}(x)=dy/dx$ is
$$
y^{\prime }=\frac{y-ut}{x}.\tag{A}
$$
Solving for $t$ we get
$$
t=\frac{y-xy^{\prime }}{u}.\tag{A'}
$$
Let $s$ be the distance traveled by the dog from $S$ to $P$, i.e. the
length of the arc $SP$ measured along $C$. Since the arc length formula is
the integral
$$
s=\int_{x}^{L}\sqrt{1+\left( f^{\prime }(\xi )\right) ^{2}}d\xi
=-\int_{L}^{x}\sqrt{1+\left( f^{\prime }(\xi )\right) ^{2}}d\xi ,
\tag{B}$$
and $s=vt$, we have
$$
t=\frac{s}{v}=-\frac{1}{v}\int_{L}^{x}\sqrt{1+\left( f^{\prime }(\xi
)\right) ^{2}}d\xi =\frac{y-xy^{\prime }}{u}.\tag{B'}
$$
Hence, equating $(A')$ to $(B')$, we get
$$
-\frac{u}{v}\int_{L}^{x}\sqrt{1+\left( f^{\prime }(\xi )\right) ^{2}}d\xi
=y-xy^{\prime }\tag{C}
$$
- Differentiate both sides and simplify
$$\begin{eqnarray*}
-\frac{u}{v}\sqrt{1+\left( y^{\prime }\right) ^{2}} &=&\frac{d}{dx}\left(
y-xy^{\prime }\right) \\
-\frac{u}{v}\sqrt{1+\left( y^{\prime }\right) ^{2}} &=&y^{\prime }-\left(
y^{\prime }+xy^{\prime \prime }\right) =-xy^{\prime \prime }.
\end{eqnarray*}$$
to get the following differential equation
$$
\sqrt{1+\left( y^{\prime }\right) ^{2}}=kxy^{\prime \prime },\qquad k=\frac{v}{u}>1.\tag{D}
$$
Set $w=y^{\prime }$ and solve $(D)$ for $w$ applying the method of separation of variables. Then
$$
\sqrt{1+w^{2}}=kxw^{\prime }=kx\frac{dw}{dx}\Leftrightarrow \frac{dw}{\sqrt{
1+w^{2}}}=\frac{dx}{kx}.\tag{E}
$$
So
$$\begin{eqnarray*}
\int \frac{dw}{\sqrt{1+w^{2}}} &=&\int \frac{dx}{kx}+C \\
\text{arcsinh }w &=&\frac{1}{k}\ln x+\ln C_{1}.\tag{F}
\end{eqnarray*}$$
The initial condition $x=L,w=y^{\prime }(L)=0$ determines the constant $C_{1}$
$$
0=\frac{1}{k}\ln L+\ln C_{1}\Rightarrow C_{1}=e^{-\frac{1}{k}\ln L}.
$$
Consequently,
$$
\text{arcsinh }w=\frac{1}{k}\ln x-\frac{1}{k}\ln L=\frac{1}{k}\ln \frac{x}{L}.\tag{G}
$$
Solve $(G)$ for $w$ and rewrite in terms of exponentials using the definition of $\sinh z=\frac{1}{2}\left( e^{z}-e^{-z}\right) $
$$
\frac{dy}{dx}=w=\sinh \left( \frac{1}{k}\ln \frac{x}{L}\right) =\frac{1}{2}\left( \left( \frac{x}{L}\right) ^{1/k}-\left( \frac{x}{L}\right)
^{-1/k}\right)\tag{H}
$$
This last differential equation is easily integrable
$$\begin{eqnarray*}
y &=&\frac{1}{2}\int \left( \frac{x}{L}\right) ^{1/k}-\left( \frac{x}{L}
\right) ^{-1/k}dx \\
&=&\frac{1}{2}\left( \frac{L}{1/k+1}\left( \frac{x}{L}\right) ^{1/k
+1}-\frac{L}{1-1/k}\left( \frac{x}{L}\right) ^{1-1/k}\right) +C
\end{eqnarray*}\tag{I}$$
Find $C$ making use of the initial condition $x=L,y=0$
$$\begin{eqnarray*}
0 &=&\frac{1}{2}\left( \frac{L}{1/k+1}\left( \frac{L}{L}\right) ^{1/k+1}-\frac{L}{1-1/k}\left( \frac{L}{L}\right) ^{1-1/k}\right) +C \\
&\Rightarrow &C=\frac{Lk}{k^{2}-1}.
\end{eqnarray*}$$
The equation of the trajectory is thus
$$
y=\dfrac{L}{2}\left( \dfrac{1}{\dfrac{1}{k}+1}\left( \dfrac{x}{L}\right) ^{\dfrac{1}{k}+1}-\dfrac{1}{1-\dfrac{1}{k}}\left( \dfrac{x}{L}\right) ^{1-\dfrac{1}{k}}\right) +\dfrac{Lk}{k^{2}-1}.\tag{J}
$$
To obtain the time $T$ the dog takes to catch the cat, make $x=0$ in the last equation and observe that the cat travels the distance $y=f(0)=uT$ (point $(R)$):
$$
y=f(0)=\frac{Lk}{k^{2}-1}=\frac{Lv/u}{\left( v/u\right) ^{2}-1}=\frac{uv}{v^{2}-u^{2}}L=uT.\tag{K}
$$
Therefore
$$
T=L\frac{v}{v^{2}-u^{2}}.\tag{L}
$$
--
References:
Pursuit Curves by Michael Lloyd
Wikipedia Entry Pursuit curve
German Wikipedia Entry Radiodrome
The Curve of Pursuit by Helmut Knaust Math 3226 Laboratory 2B
ADDED. Let $M$ be the point $(L/2,0)$. We can easily verify that the total length of $C$ is equal to $\overline{SM}+\overline{MR}$.