Does $mathcal{A}otimesmathbb{C}(t)congmathcal{D}otimesmathbb{C}(t)$ imply an isomorphism of Lie algebras?

Put $mathcal{A}'=mathcal{A}otimes_{mathbb{C}}mathbb{C}(t)$ and similarly for $mathcal{D}'$. Choose an isomorphism $fcolonmathcal{D}'tomathcal{A'}$.
Choose a (countable) basis $mathcal{D}_0={d_i:iinmathbb{N}}$ for $mathcal{D}$ over $mathbb{C}$. Then $f(mathcal{D}_0)$ is a basis for $mathcal{A}'$ over $mathbb{C}(t)$ but $dim_{mathbb{C}}(mathcal{A}')=dim_{mathbb{C}}(mathcal{A})$ so we can also choose a countable basis $mathcal{A}_0={a_i:iinmathbb{N}}$ for $mathcal{A}$ over $mathcal{C}$. We must have $[d_i,d_j]=sum_kp_{ijk}d_k$ and $[a_i,a_j]=sum_kq_{ijk}a_k$ for some structure constants $p_{ijk},q_{ijk}inmathbb{C}$. Now let $K$ be a countable subfield of $mathbb{C}$ containing all these structure constants, and also all the constants needed to ensure that $f(mathcal{A}_0)subseteq K(t).mathcal{D}_0$ and $f^{-1}(mathcal{D}_0)subseteq K(t).mathcal{A}_0$. Put $mathcal{A}_1=K.mathcal{A}_0$ and $mathcal{D}_1=K.mathcal{D}_0$, so these are Lie algebras over $K$ that become isomorphic over $K(t)$. If $mathbb{C}$ were algebraic over $K$ then it would be countable, which is false. Thus, we can choose an embedding $icolon K(t)tomathbb{C}$ extending the identity on $K$. By applying this to the coefficients of $f$, we obtain an isomorphism $mathcal{D}simeqmathcal{A}$.