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 $f(j)<f(i)$.

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So let's assume that such a function $f$ existed. We find the smallest value $N \in \mathbb{N}$ such that we have $f(N)=a_{n_1}+b_{m_1}$ with ordinal $n_1<\alpha$ satisfying the following property: "for all natural numbers $i > N$, whenever we decompose $f(i)$ into $a_{k_1}+b_{k_2}$ [with $k_1<\alpha$ and $k_2<\beta$] we must have $k_1 \geq n_1$." 

So essentially, I think we just want to look at how the decomposition of $f(x)$ would occur for values $x \geq N$. 
It seems that we want to show that there exists a smallest value $N_1 \in \mathbb{N}$ with $N_1 \geq N$ such that if $f(N_1)=a_{n_1}+b_{z}$ (where $z<\beta$), then the following property is satisfied: "For all natural numbers $i > N_1$, whenever we decompose $f(i)$ into $a_{k_1}+b_{k_2}$ [with $k_1<\alpha$ and $k_2<\beta$] we must have $k_1 > n_1$." 

But now if we write $f(N_1+1)=a_{n_2}+b_{m_2}$, then because we have $n_2>n_1$, we must get $m_2<m_1$. And now this is suggestive of defining a smallest value $N_2 \geq N_1+1$ such that if $f(N_2)=a_{n_2}+b_{z}$, then the following property is satisfied: "For all natural numbers $i > N_2$, whenever we decompose $f(i)$ into $a_{k_1}+b_{k_2}$ [with $k_1<\alpha$ and $k_2<\beta$] we must have $k_1 > n_2$."  

But now if we write $f(N_2+1)=a_{n_3}+b_{m_3}$, then because we have $n_3>n_2$, we must get $m_3<m_2$. And this is suggestive of defining $N_3 \geq N_2+1$ such that we can write $f(N_3+1)=a_{n_4}+b_{m_4}$. Similarly because we have $n_4>n_3$, we must get $m_4<m_3$.

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Once we fully generalize the argument in the previous part of the answer, we get impossibility of such a function  $f$ due to a lack of infinite chain of the form (note that comparisons in this chain are ordinal comparisons): $......<m_5<m_4<m_3<m_2<m_1$.