Here is a stab at an answer, but it is incomplete.

Let $S:={\rm Spec}\, B$ and let $\widehat{S}:={\rm Spec}\,\widehat{B}$ be the completion of
$B$ along its maximal ideal $m$. Let $\phi:\widehat{S}\to S$ be the natural morphism (which is faithfully flat).
The composition of morphisms $\widehat{S}\to S\to{\rm Spec}\,k$ gives rise to a triangle
of cotangent complexes and hence to an exact sequence
$$
\dots\to T^1_{\phi}\to T^1_{\widehat{S}/k}\to \phi^*T^1_{S/k}\to T^2_{\phi}\to T^2_{\widehat{S}/k}\to \phi^*T^2_{S/k}\to T^3_\phi\to\dots {\rm (*)}
$$
so what you need to show is that $T^i_\phi$ vanishes for $i=1,2,3$. You would then get
isomorphisms $T^i_{\widehat{S}/k}\to \phi^*T^i_{S/k}$ and since you can repeat this for $B'$ instead of $B$, you would
get the required isomorphism.

Let $L_\phi$ be the cotangent complex of $\widehat{S}/S$. This complex
$L_\phi$ is concentrated in degree $0$. A quick way to see this is to notice that

the formation of the cotangent complex is compatible with direct limits of rings (see
Quillen, "Cohomology of commutative rings", eq. (4.11))

the ring $B$ is excellent because it is essentially of finite type over a field (Grothendieck) and thus the fibres of $\phi$ are geometrically regular;

a (deep...) result of Popescu (see for instance Th. 1.3 in "Approximations of versal deformations", by B. Conrad
and AJ de Jong) then implies that $\widehat{B}$ is a direct limit of
smooth $B$-algebras, and for the latter the cotangent complex is clearly concentrated at $0$.

By considering the sequence analogous to (*) for
the modules $T_i$ instead of $T^i$, this proves the analog of your assertion for the $T_i$. Furthermore, we see that
$T^i_\phi={\rm Ext}^i(L_\phi,\widehat{B})={\rm Ext}^i(\Omega_\phi,\widehat{B})$.

So the issue is to show that ${\rm Ext}^i(\Omega_\phi,\widehat{B})=0$ for $i>0$. Maybe your hypothesis on the fact that the singularity of $S$ is isolated at the closed point plays a role here.