I'm late to the party with this answer, but the question is on the front page, so whatever.

As David essentially said, given any $\mathrm{U}(1)$-bundle/connection pair over $M$ with connection $\alpha$, one can obtain any other bundle/connection pair by tensoring with a flat bundle over $M$. Flat bundles are characterized by $H^1(M, \mathrm{U}(1))$, which is isomorphic to the character group of $\pi_1(M)$ (for $M$ path-connected at least). One way to see this, which is less powerful than David's characterization, but a bit easier to grasp, is that a flat connection on a (right) $G$-bundle $P$ is equivalent to an involutive, hence integrable, distribution on $P$. We can lift any curve starting at $m\in M$ to a horizontal curve through any $p\in\pi^{-1}(m)$, where $\pi:P\rightarrow M$ is the bundle projection. The maximal integral manifold of the distribution through $p$ is actually a covering space of $M$ via $\pi$, and so it follows that lifts through $p$ of homotopic curves in $M$ are homotopic in $P$, and have the same endpoint. Call this endpoint $E_p([\gamma])$, where $[\gamma]$ is the homotopic class of the curve $\gamma$. If $\gamma$ is a *closed* curve, then the endpoint of its lift lies in the same fiber as $p$, and we can write
$$
E_p([\gamma]) = p\cdot\chi_p([\gamma]^{-1})
$$
where $\chi_p:\pi_1(M)\rightarrow G$. It is not difficult to show that $\chi_p$ is a homomorphism. $P$ can then be reconstructed as the quotient of the trivial bundle $\widetilde{M}\times G$ with the obvious flat connection under the equivalence relation
$$
([\delta],g) \sim (\[\gamma]*[\delta],\chi_p([\gamma])g)
$$
by mapping $\left[[\delta],g\right]_\sim\mapsto E_p([\delta])\cdot g$. Here $\widetilde{M}$ is the universal covering space of $M$, thought of as classes of homotopic curves starting at $m$.
Hence all such flat bundles are characterized by the set of homomorphisms $\chi_p:\pi_1(M)\rightarrow G$, which in the case $G=\mathrm{U}(1)$ is just the character group of $\pi_1(M)$. (It's very easy to screw up the conventions here; hopefully I haven't.)

To answer your second question, it depends what you mean by "curvature in the line bundle". Normally one would think of the curvature $\alpha$ as a $\mathfrak{g}$-valued two-form on $P$. Maybe this is what you're getting at: given a representation $\lambda:G\rightarrow \mathrm{GL}(V)$ on the vector space $V$, we form the vector bundle $P\times_\lambda V$ associated to $P$. Similarly we can form the associated bundle $P\times_{\mathrm{Ad}}~\mathfrak{g}$, where $\mathrm{Ad}:G\rightarrow \mathrm{GL}(\mathfrak{g})$ is the adjoint representation. The infinitesimal $\mathfrak{g}$-action $\lambda'(\xi)v = \frac{d}{dt}\lambda(\exp(\xi t))v\big\vert_{t=0}$ on $V$ induces a well-defined action of $P\times_{\mathrm{Ad}}~\mathfrak{g}$ on $P\times_\lambda V$.

Using the curvature $\alpha$ we define a $P\times_{\mathrm{Ad}}~\mathfrak{g}$-valued 2-form on $M$, given by
$$
\beta_m(X_m,Y_m) = [p,\alpha_p(A_p,B_p)]_{\mathrm{Ad}}
$$
where $A_p, B_p$ are (arbitrary) lifts of $X_m, Y_m$ to $p\in \pi^{-1}(m)$. The fact that $\alpha$ vanishes on vertical vectors, coupled with the $G$-equivariance of $\alpha$, ensures that $\beta$ is well-defined.

It's possible (and a worthwhile exercise) to prove that
$$
([\nabla_X,\nabla_Y]-\nabla_{[X,Y]})s = \beta(X,Y)\cdot s
$$
where $s$ is a section of the associated vector bundle $P\times_\lambda V$. (The left hand side appears to depend on the vector fields $X,Y$, but in fact at a particular point it just depends on their values at that point.)

If one expresses $s$ in terms of its corresponding $G$-equivariant function $\widetilde{s}:P\rightarrow V$, i.e.
$$
s(m) = [p,\widetilde{s}(p)]_\lambda \qquad\textrm{any }p\in\pi^{-1}(m)
$$
then this unpacks to
$$
\left[p, ([X^h,Y^h]-[X,Y]^h)_p \widetilde{s}(p)\right] _
\lambda = \left[p, \lambda'(\alpha_p(X^h_p,Y^h_p)) \widetilde{s}(p)\right]_\lambda
$$
($X^h$ being the horizontal lift of $X$ etc.)

As with the curvature $\alpha$, in the case when $G$ is abelian you don't need to go to all this trouble, and can just define $\beta$ as a $\mathfrak{g}$-valued 2-form on $M$. Then as you guessed, these 2-forms are related by $\beta=\lambda'\circ\alpha$.