Let $C_0$ be the plane affine curve given by the equation $$ f = y^2 - x (x - 1) (x - \lambda),$$
$ R = k[x,y] / (f)$, and $S$ be a ring isomorphic to $R$.
From one side, we can reconstruct the projective completion of $C_0$, (let us call it $C$), from $S$; and hence, if not $\lambda$, at least the $j$-invariant of the curve $C$, $j(\lambda)$. (One of the constructions would be to construct the field of fractions $K = \mathbb{Q}(R)$, choose any transcendent element $t$, consider the field extension $K / k(t)$, choose the primitive element/polynomial of $K$ over $k(t)$; then consider the "second chart" $u = 1/t$, and so on.)
From another point of view, there are no deformations of the ring $S$ (as an abstract commutative ring without fixing an epimorphism $ k[x,y] \rightarrow S$), as explained, for example, in Michael Artin's lectures on deformation theory [1].
Why there is no contradiction here?
Of course, the question stands for any smooth plane affine curve of degree at least 3.
In other words, the functor of deformations "does not commute with completions". Can someone elucidate, and/or explain this? -- Thank you.
Part (b) to the same question: what is the "explanation" why $H^1(C_0,T)$ (which is zero) does not compute the tangent space to this moduli space of affine elliptic curves, $A^1$ with coordinate $j$, (or one with the $\lambda$-coordinate? )
Thank you.
[1] Michael Artin, "Deformation theory", Course Notes, Fall 1996, page 3 (Corollary 1.7).
There is no contradiction here because the theorem that you state (which is a theorem that shows right up at the start in any deformation theory lesson) is a theorem for local deformations. This does not give you any information about global families.
More precisely, we restrict to $ \lambda \neq 0,1 $ to see that there is a 1-parameter family of affine plane smooth curves given by $$ \operatorname{Spec}\mathbb{C}[\lambda][x,y]/y^2 - x(x-1)(x-\lambda)) \rightarrow \operatorname{Spec}C[\lambda] - \{ 0,1 \} $$
Fix some point $ \mu \in \mathbb{C} - \{ 0,1 \} $. The theorem says that since the ring $ R_{\mu} = \operatorname{Spec}\mathbb{C}[x,y]/(y^2 - x(x-1)(x-\mu)) $ is smooth, it has no formal deformations. That is to say if $ A $ is an Artin local $ \mathbb{C} $-algebra and $ S $ is an $ A $-flat algebra such that $ S \otimes_A \mathbb{C} \cong R_{\mu} $, then $ S $ is isomorphic to the trivial deformation $ R_{\mu} \otimes_{\mathbb{C}} A $. Hence the family is formally trivial around $ \mu $ but it isn't so Zariski locally around $ \mu $ as I'm sure you know - the $ j $-invariant of $ R_{\mu} $ remains the same only for the six cross ratios of $ \mu $. Here I took the field to be $ \mathbb{C} $ to avoid issues of characteristic.