Yes. In general you only have subadditivity of the limit inferior (that is, in general you only have $\liminf (a_n+b_n) \geq \liminf(a_n) + \liminf(b_n)$) but if at least one sequence converges, then you have equality.
To see why you have equality, think of $\liminf$ as the infimum of all limits of converging subsequences. If $a_{j_k}+b_{j_k}$ is a converging subsequence of $\{a_n+b_n\}$, then $b_{j_k}$ must also converge (since $a_{j_k}$ converges: every subsequence of a converging sequence converges); and likewise, if $b_{j_k}$ is a converging subsequence of $b_n$, then $a_{j_k}+b_{j_k}$ is a converging subsequence of $a_n+b_n$. Moreover, if $b_{j_k}$ converges to $s$, then $a_{j_k}+b_{j_k}$ converges to $a+s$. Conversely, if $a_{j_k} + b_{j_k}$ converges to $a+t$, then $b_{j_k}$ must converge to $t$. So if $L$ is the set of all points that are limits of subsequences of $b_n$, then the set of all points that are limits of subsequences of $a_n+b_n$ is $a+L = \{a+t \mid t\in L\}$. Therefore
\[ \liminf (a_n+b_n) = \inf (a+L) = a + \inf L = a + \liminf b_n = \lim_{n\to\infty}a_n + \liminf b_n.\]
To get the difference, just apply this to $a_n$ and $(-b_n)$.
(Or you can try the same argument with $\limsup$, and use the fact that $\liminf -b_n = -\limsup b_n$; you would think of $\limsup$ as the supremum of all limits of converging subsequences of $b_n$).
Now see if you can figure out why you only have an inequality in the general case...