I suspect that probably this is identical (or extremely similar) to nombre's answer. But I was having some difficulty understanding it. So perhaps someone else might find this useful (in part, I am writing to better understand this myself).
Let's suppose the well-order of $A$ and $B$ have order-types $\alpha$ and $\beta$ respectively. So we use the notation $a_i$ ($i<\alpha$) to denote the $i$-th element of $A$. We have $a_i<a_j$ whenever $i<j<\alpha$. Similarly, we use $b_i$ ($i<\beta$) to denote the $i$-th element of $B$.
Since $A+B$ is already linearly-ordered, we want to prove that $A+B$ has no infinite descent. In other words, we want to disprove the existence of a function $f:\mathbb{N} \rightarrow \mathbb{R}$ such that (i) For all $x$ in domain of $f$ we have $f(x)=a_i+b_j$ (for some $i<\alpha$ and $j<\beta$) (ii) $f$ is a 1-1 function (iii) For all $i,j \in \mathbb{N}$ (with $j>i$) we must have $f(j)<f(i)$.