I think the 2 algebras $L_{0}$ and $L_{1}$ are isomorphic to the Lie algebra $\mathfrak{iso}(2)$ .
I struggled to find the isomorphism $\phi$ between the Lie algebra $L_{0}$ and $L_{1}$ ?
Let $L_{0}=Span\{v_{1},v_{2},v_{3},v_{4}\}$ and $L_{1}=Span\{w_{1},w_{2},w_{3},w_{4}\}$ the Lie algebras.
such that
$ v_{1}=\frac{\partial}{\partial t}\quad,\quad v_{2}=u\frac{\partial}{\partial u}\quad,\quad v_{3}=\sqrt{x}\exp(\frac{b}{2}t)\frac{\partial}{\partial x}+2b\exp(\frac{b}{2}t)\sqrt{x}u\frac{\partial}{\partial u}\quad,\quad v_{4}=\sqrt{x}\exp(-\frac{b}{2}t)\frac{\partial}{\partial x}$ $ w_{1}=\frac{\partial}{\partial t}\quad,\quad w_{2}=u\frac{\partial}{\partial u}\quad,\quad w_{3}=\sqrt{y}\exp(\frac{e}{2}t)\frac{\partial}{\partial y}+2e\exp(\frac{e}{2}t)\sqrt{y}u\frac{\partial}{\partial u}\quad,\quad w_{4}=\sqrt{y}\exp(-\frac{e}{2}t)\frac{\partial}{\partial y}$
with $\begin{array}{c|c|c|c|c|} & v_{1} &v_{2} & v_{3}& v_{4} \\ \hline v_{1} & 0 &0&\frac{b}{2}v_{3}&-\frac{b}{2}v_{4} \\ \hline v_{2} &0 &0&0&0 \\ \hline v_{3} &-\frac{b}{2}v_{3} &0 &0&-bv_{2} \\ \hline v_{4} &\frac{b}{2}v_{4} &0 &bv_{2}&0\\ \hline \end{array}$ and $ \begin{array}{c|c|c|c|c|} & v_{1} &v_{2} & v_{3}& v_{4} \\ \hline v_{1} & 0 &0&\frac{e}{2}v_{3}&-\frac{e}{2}v_{4} \\ \hline v_{2} &0 &0&0&0 \\ \hline v_{3} &-\frac{e}{2}v_{3} &0 &0&-ev_{2} \\ \hline v_{4} &\frac{e}{2}v_{4} &0 &ev_{2}&0\\ \hline \end{array}$ are Lie commutators. Such that $\boxed{b\ne\pm e}$.
We can rewrite the condition that $f\colon L_0\rightarrow L_1$ is a bijective Lie algebra homomorphism by saying that $f([x,y]_{L_0})=[f(x),f(y)]_{L_1}$ for all $x,y\in L_0$. In operator form this means $f\circ ad(x)=Ad(f(x))\circ f$ for all $x\in L_0$, so that $$ ad(x)=f^{-1}\circ Ad(f(x))\circ f. $$ It is enough to consider a basis for $L_0$ for the values of $x$. The characteristic polynomials of $ad(e_i)$ and $Ad(f(e_i))$ must coincide. This gives equations in the variables of $f$. Now a direct computation of all possible $f$ is easy (note that $\det(f)\neq 0$ is a strong condition as well). For example, we immediately obtain $b=e\cdot f_{11}$ for nonzero $b,e$. So isomorphisms are given, for example, by the diagonal matrices $$ f=\operatorname{diag}(f_{11},\frac{f_{33}f_{44}}{f_{11}},f_{33},f_{44}) $$ for arbitrary nonzero diagonal elements $f_{33},f_{44}$ from the field, where $f_{11}=\frac{b}{e}$. Note that $e$ is not necessarily $\exp(1)$.