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1. About the second definition:
   $\alpha$) It is not true that for an arbitrary ring a) is equivalent to c):
   Indeed Bourbaki in Algèbre commutative, Chapitre II, Exercices §5, 12) c) exhibits a ring $B$ and a projective module of rank $1$ over $B$ which is not isomorphic to an invertible fractional ideal of $B$. This does not contradict Eisenbud's Theorem 11.6 because $R$ is explicitly supposed noetherian there.
   $\beta$) It is also not true that b) is equivalent to c):
   Take $R=\mathbb Z$ and $\mathbb Z\subsetneq M=\bigcup \frac {1}{p_1\cdots p_i}\mathbb Z\subsetneq \mathbb Q$ where $p_i$ is the $i$-th prime.
   Then for all prime $\mathfrak p\subset \mathbb Z$ the $\mathbb Z_\mathfrak p$-module $M_\frak p$ is free of rank $1$, since $$M_{(0)}=\mathbb Z_{(0)}(=\mathbb Q) \quad \operatorname {and}\quad M_{(p_i)}=\frac {1}{p_i}\mathbb Z_{(p_i)}$$ However the $\mathbb Z$-module $M$ is neither finitely generated nor projective (over $\mathbb Z$, projective=free)

2. I think the only reasonable definition of $\operatorname {Pic(R)}$ valid for any commutative ring is to define it as the Picard group of the affine scheme $X=\operatorname {Spec}(R)$.
   As is the case for any locally ringed space $(X,\mathcal O_X)$ the Picard group consists of isomorphism classes of locally free $\mathcal O_X$-Modules of rank one.
   This is exactly the definition used with much success for general, non-affine, schemes but also for topological spaces, differential manifolds, etc.

3. In our special case $X=\operatorname {Spec}(R)$ the definition in 2) translates in purely algebraic terms to:
   $\operatorname {Pic(R)}$ consists of isomorphism classes of $R$-modules $M$ such that there exist finitely many elements $f_1,\cdots,f_n\in R$ with:
   $\alpha$) $\sum Rf_i=R$
   $\beta$) $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.
   (Remark: $\beta$) implies that $M$ is finitely generated over $R$)

4. The modules $M$ in 3) can also be characterized as the finitely generated modules over $R$ such that equivalently:
   i) $M$ is projective of rank $1$
   ii) The modules $M_\frak m$ are free of rank $1$ over $R_\frak m$ for all maximal ideals $\mathfrak m\subset R$
   iii) The canonical $R$-linear map $M\otimes_RM^*\to R:m\otimes \phi\mapsto\phi(m)$ is bijective
   Note that these are pleasant algebraic characterizations of the modules generating $\operatorname {Pic(R)}$, but the better, conceptual definition is that given in 2) and 3).
   Caveat: As shown in 1) above, the hypothesis that $M$ is finitely generated is vital.

Edit: WARNING
The confusion is made worse by Bourbaki's unfortunate decision to define a projective module of rank $1$ as a finitely generated module $P$ for which $M_\mathfrak p$ is free of rank $1$ over $R_\mathfrak p$ for all primes $\mathfrak p\subset R$.
As my example 1) $\beta$) shows, omitting to require that $P$ be finitely generated [as is done in Pete's condition (2) b)] means accepting modules which aren't even projective, and which don't satisfy (2) a) nor (2) c) of the question.