Yes, it is true, though an algebraic proof seems (there may be a simpler proof however) somewhat tricky.

- Such a line bundle lies in $\mathrm{Pic}^\tau(X)$. This is a general fact as a line bundle lies in $\mathrm{Pic}^\tau(X)$ if its rational Chern classes are trivial (this follows from Riemann-Roch) and the Chern classes of a line bundle with integrable connection are torsion (an algebraic proof is given in Dix exposés sur la cohomologie des schémas). This works without the assumption of $X$ being abelian.
- It is then a further fact that for abelian varieties $X$ we have that $\mathrm{Pic}^\tau(X)=
\mathrm{Pic}^0(X)$ (this I think is an Mumford's abelian varieties somewhere).

This extends to immediately to families by checking fibre by fibre.

**Addendum**: Sorry forgot to say that this all requires characteristic zero. In characteristic $p$ every $p$'th power line bundle has an integrable connection (even of $p$-curvature $0$) but will in general not lie in $\mathrm{Pic}^0(X)$.

**Addendum 1**: Lars (in a comment) makes an interesting point about the positive characteristic situation. A module structure over the ring of differential operators (aka a stratification) implies in particular that the line bundle is a $p^n$'th power for each $n$ and as $\mathrm{Pic}(X)/\mathrm{Pic}^\tau(X)$ is a finitely generated group this implies that the line bundle lies in $\mathrm{Pic}^\tau(X)$. The same idea could be applied to the characteristic zero situation if the $p$-curvature of the reduction modulo $p$ for an infinite number of primes $p$ were zero. However that should be true only if the line bundle has finite order so it doesn't give very much.

**Addendum 2**: Veen is asking about the equality $\mathrm{Pic}^0(A)=\mathrm{Pic}^\tau(A)$ particularly in a family (when $A$ is abelian). The easiest way to answer all of these questions simultaneously is to assume that there is an ample line bundle $\mathcal L$ on $A$ (which is true locally on the base) and then consider the map $A\to \mathrm{Pic}(A)$ given by $a\mapsto \mathcal L_a\bigotimes\mathcal L^{-1}$. To get all equalities needed it is enough to show that the image is $\mathrm{Pic}^\tau(A)$. This is something that can be checked fibrewise and then it can be extracted from Mumford.