have you solution?

$a_n=b_n-b_{n-1}-h_n,h_n\in \{0,1,2\}.$

Rust wrote: $a_{n}=b_{n}-b_{n-1}-h_{n},h_{n}\in \{0,1,2\}.$ I'm sorry, but how?

Dear Rust,what are you talking about?

Yes, there always exists a sequence $\{a_n\}$ satisfying the requirements. The idea is to choose consecutively the terms of $a_n$ watching out the sums $a_i+a_j$ to avoid any $b_r$. First, we take some $b_r, b_r>b+1$ and set $a_1 := b_r$ if $b_{r+1}\neq 2b_r$ and $a_1 :=b_r-1$ otherwise. Suppose now, we have chosen all $a_i$'s up to some $b_n, n\geq 1$, that's we have constructed $a_1<a_2<\dots<a_k$ such that: 1) $a_k < b_n, b_n-a_k\leq a$ 2) $a_{i+1}-a_i\in \{a,b\}, i=1,2,\dots,k-1$ 3) $a_i+a_j\not\in \{b_m\}_{m=1}^{\infty}\,,\,1\leq i,j\leq k$. We want to define the next $a_i$'s up to $b_{n+1}$, that is $a_{k+1}<a_{k+2}<\dots<a_{\ell}<b_{n+1}\,,\,b_{n+1}-a_{\ell}\leq a$. Since $b_{n+1}\geq 2b_n, b_{n+2}\geq 2b_{n+1}$, we only need to ensure $a_i+a_j\neq b_{n+1}\,,\,i=k+1,k+2,\dots,\ell\,,\, 1\leq j\leq i$. Let's begin with $a_{k+1}$. The two possibilities for $a_{k+1}$ are $a_{k+1}=a_k+a$ or $a_{k+1}=a_k+b$. Suppose, there exists $a_j, 1\leq j\leq k+1$ with $(a_{k}+a)+a_j=b_{n+1}$. In that case there does not exists $a_r, 1\leq r\leq k+1$ with $(a_{k}+b)+a_r=b_{n+1}$. Indeed, assuming it, we get $a_j-a_r=b-a$, which would imply $a\mid b$, a contradiction. Thereby, we can safely choose one of the two options $a_{k+1}=a_k+a$ or $a_{k+1}=a_k+b$. Applying the same arguments, we can choose consecutively $a_{k+2}, a_{k+3},\dots$, till we come to $a_{r}$, for which $b_{n+1}-a_{r}\leq b$. Then we add to $a_r$ several times $a$, till we get to $a_{\ell}$ for which $a_{\ell} < b_{n+1}, b_{n+1}-a_{\ell}\leq a$ In that way we have extended $a_i$'s up to the next $b_{n+1}$, still complying the requirements. Thus, step after step, we can construct the sequence $\{a_n\}_{n=1}^\infty$.