In both cases $A$ and $A^\wedge$ are reduced. In case (2) because the completion of a reduced local Nagata ring is reduced (More on Algebra, Lemma \ref{more-algebra-lemma-completion-reduced}). In both cases $A$ and $A^\wedge$ have dimension $1$ (More on Algebra, Lemma \ref{more-algebra-lemma-completion-dimension}). The $\delta$-invariant and the number of geometric branches of $A$ and $A^\wedge$ agree by Varieties, Lemma \ref{varieties-lemma-delta-same-after-completion} and More on Algebra, Lemma \ref{more-algebra-lemma-one-dimensional-number-of-branches}. Let $A'$ be the integral closure of $A$ in its total ring of fractions as in Varieties, Lemma \ref{varieties-lemma-pre-delta-invariant}. By Varieties, Lemma \ref{varieties-lemma-normalization-same-after-completion} we see that $A' \otimes_A A^\wedge$ plays the same role for $A^\wedge$. Thus we may replace $A$ by $A^\wedge$ and assume $A$ is complete. \medskip\noindent Assume (1) holds. It suffices to show that $A$ has two geometric branches and $\delta$-invariant $1$. We may assume $A = \kappa[[x, y]]/(ax^2 + bxy + cy^2)$ with $q = ax^2 + bxy + cy^2$ nondegenerate. There are two cases. \medskip\noindent Case I: $q$ splits over $\kappa$. In this case we may after changing coordinates assume that $q = xy$. Then we see that $$ A' = \kappa[[x, y]]/(x) \times \kappa[[x, y]]/(y) $$ \medskip\noindent Case II: $q$ does not split. In this case $c \not = 0$ and nondegenerate means $b^2 - 4ac \not = 0$. Hence $\kappa' = \kappa[t]/(a + bt + ct^2)$ is a degree $2$ separable extension of $\kappa$. Then $t = y/x$ is integral over $A$ and we conclude that $$ A' = \kappa'[[x]] $$ with $y$ mapping to $tx$ on the right hand side. \medskip\noindent In both cases one verifies by hand that the $\delta$-invariant is $1$ and the number of geometric branches is $2$. In this way we see that (1) implies (2). Moreover we conclude that the final statement of the lemma holds. \medskip\noindent Assume (2) holds.  More on Algebra, Lemma \ref{more-algebra-lemma-number-of-branches-1} implies $A'$ either has two maximal ideals or $A'$ has one maximal ideal and $[\kappa(\mathfrak m') : \kappa]_s = 2$. \medskip\noindent Case I: $A'$ has two maximal ideals $\mathfrak m'_1$, $\mathfrak m'_2$ with residue fields $\kappa_1$, $\kappa_2$. Since the $\delta$-invariant is the length of $A'/A$ and since there is a surjection $A'/A \to (\kappa_1 \times \kappa_2)/\kappa$ we see that $\kappa = \kappa_1 = \kappa_2$. Since $A$ is complete (and henselian by Algebra, Lemma \ref{algebra-lemma-complete-henselian}) and $A'$ is finite over $A$ we see that $A' = A_1 \times A_2$ (by Algebra, Lemma \ref{algebra-lemma-finite-over-henselian}). Since $A'$ is a normal ring it follows that $A_1$ and $A_2$ are discrete valuation rings. Hence $A_1$ and $A_2$ are isomorphic to $\kappa[[t]]$ (as $k$-algebras) by More on Algebra, Lemma \ref{more-algebra-lemma-power-series-over-residue-field}. Since the $\delta$-invariant is $1$ we conclude that $A$ is the wedge of $A_1$ and $A_2$ (Varieties, Definition \ref{varieties-definition-wedge}). It follows easily that $A \cong \kappa[[x, y]]/(xy)$. \medskip\noindent Case II: $A'$ has a single maximal ideal $\mathfrak m'$ with residue field $\kappa'$ and $[\kappa' : \kappa]_s = 2$. Arguing exactly as in Case I we see that $[\kappa' : \kappa] = 2$ and $\kappa'$ is separable over $\kappa$. Since $A'$ is normal we see that $A'$ is isomorphic to $\kappa'[[t]]$ (see reference above). Since $A'/A$ has length $1$ we conclude that $$ A = \{f \in \kappa'[[t]] \mid f(0) \in \kappa\} $$ Then a simple computation shows that $A$ as in case (1).