As was noted in the comment, $d$ must be $0$ (bundles with a flat connection can have only degree $0$), and different connections can lead to the same bundle. However, every line bundle of degree zero on an elliptic (or higher genus) curve has a unique flat unitary connection, i.e. in your example with $|\lambda|=|\mu|=1$. This identifies the Jacobian of any smooth projective complex curve with a product of $2g$ circles (once you fix a basis of $H_1(C,\Bbb Z)$). For an elliptic curve, this is the usual identification with the 2-torus.
$\require{AMScd}$
The following result holds in much more generality than asked by the OP. There is no characteristic zero/smoothness assumption.
Proposition: Let $X/k$ be a qcqs scheme with $k$ separably closed. Let $K/k$ be a separable extension with $K$ separably closed. Let $n \in \mathbf{N}$ be a positive integer that is prime to the characteristic of $k$. Then the natural map
$$\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} \to \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)}$$
is an isomorphism.
The proof goes as follows. The Kummer sequence gives the following commutative diagram with exact rows:
$\require{AMScd}$
\begin{CD}
0 @>{}>>\frac{\operatorname{Pic}(X)}{n\operatorname{Pic}(X)} @>{}>> H^2(X, \mu_n)@>{}>> H^2(X,\mathbf{G}_m) \\
{} @VVV @VVV @VVV {} \\
0 @>{}>> \frac{\operatorname{Pic}(X_K)}{n\operatorname{Pic}(X_K)} @>{}>> H^2(X_K, \mu_n) @>{}>> H^2(X_K, \mathbf{G}_m)
\end{CD}
The middle vertical arrow is an isomorphism by the smooth base change theorem in \'{e}tale cohomology (Corollary 1.3.1 of this):
Smooth Base Change: Suppose $K/k$ is an extension of separably closed fields, $X$ a qcqs $k$-scheme and $\mathscr{F}$ a torsion sheaf with torsion orders not divisible by the characteristic of $k$.Then, the base change map
$$H^i(X,\mathscr{F}) \to H^i(X_K, \mathscr{F}_K)$$
is an isomorphism for all $i \geq 0$.
Therefore to prove the proposition, we reduce to proving the following injectivity statement concerning Brauer groups:
Proposition: Let $k \subseteq K$ be a separable extension of fields with $k$ separably closed. Let $X/k$ be a qcqs scheme. Then $H^2(X,\mathbf{G}_m)$ injects into $H^2(X_K, \mathbf{G}_m)$.
To prove this, write $K$ as a direct limit of smooth $k$-algebras. Since cohomology commutes with direct limits for qcqs schemes, if a (cohomological) Brauer class $\alpha$ in $H^2(X,\mathbf{G}_m)$ dies in $H^2(X_K,\mathbf{G}_m)$ it must die in $H^2(X_R, \mathbf{G}_m)$ for some smooth $k$-algebra $R$. By smoothness and using that $k$ is separably closed, $\operatorname{Spec} R$ has a $k$-rational point. Therefore the canonical map $f: X_{R} \to X$ has a section $g$ and so on cohomology the composition
$$H^2(X,\mathbf{G}_m) \stackrel{f^\ast}{\to} H^2(X_R,\mathbf{G}_m) \stackrel{g^\ast}{\to} H^2(X,\mathbf{G}_m)$$ is the identity. We conclude that $\alpha = 0$ as desired.
Finally we remark that in the injectivity statement on Brauer groups, only the ground field is required to be separably closed. There is no requirement on $K$, only that $K/k$ be separable.
Edit: The proof also shows that for any \'{e}tale sheaf $\mathscr{F}$ on $X$ such that multiplication by $n$ is surjectve, then $$\frac{H^i(X,\mathscr{F})}{nH^i(X,\mathscr{F})} \to \frac{H^i(X_K,\mathscr{F})}{nH^i(X_K,\mathscr{F})} $$
is an isomorphism for all $i \geq 1$.
Best Answer
Let me address the last part of your question.
Let $X$ be a smooth, projective variety over an arbitrary ground field $k$.
I want to write $Pic^{[L]}(X)$ instead of $Pic^L(X)$ -- i.e., to make explicit that the variety depends only on the Neron-Severi class of $L$ -- for reasons which will become clear shortly.
Suppose first that $L$ is algebraically equivalent to $0$. Then $Pic^{[L]}(X) = Pic^0(X)$, so certainly it is an abelian variety.
Next suppose that $L$ is a $k$-rational line bundle on $X$. Then $Pic^{[L]}(X)$ is not literally an abelian variety, because it is a nonidentity coset of a group rather than a group itself. However, it is canonically isomorphic to the abelian variety $Pic^0(X)$ just by mapping a line bundle $M$ to $M - L$. So it might as well be an abelian variety, really.
Finally, supose that $L$ is not itself $k$-rational but that its Neron-Severi class $L$ is rational -- i.e., $L$ is given by a line bundle over the algebraic closure which is algebraically equivalent to each of its Galois conjugates. Then $Pic^{[L]}(X)$ is a well-defined principal homogenous space of the Picard variety $Pic^0(X)$ but need not have any $k$-rational points. For instance, suppose that $X$ is a curve. Then the Galois action on the Neron-Severi group is trivial, so taking $L/\overline{k}$ to be any degree $n$ line bundle, we get $Pic^{[L]}(X) = Pic^n(X) = Alb^n(X)$, a torsor whose $k$-rational points parameterize $k$-rational divisor classes of degree $n$. (Note that here when I write $Pic^0(X)$ I am talking about the Picard variety rather than the degree $0$ part of the Picard group. More careful notation would be $\underline{\operatorname{Pic}}^0(X)$.)
In particular, if $X$ is a genus one curve, then there is a canonical isomorphism $X \cong Pic^1(X)$, so $Pic^1(X)$ can be endowed with the structure of an abelian variety iff $X$ has a $k$-rational point.
Some further material along these lines can be found in Section 4 of
http://alpha.math.uga.edu/~pete/wc2.pdf