Suppose I have a sequence $a_n$, whose entries are the ordered elements of $S_{x,y}$:
$S_{x, y}= \{ z \mid \left( z=n_1x+n_2y \right) \wedge \left( n_1, n_2 \in \mathbb{N}_1 \right) \wedge \left( \gcd \left( n_1, n_2 \right) = 1 \right)\}$
$x, y \in \mathbb{N}_1$
The Question:
Given that $\gcd \left( x, y \right) = 1$, is it true that $a_n - a_{n-1} = 1$ for sufficiently large $n$?
In Layman's Terms:
This is nearly equivalent to the coin problem, except that my two "coins" can only be added in ways where the amounts of each "coin" are coprime (in addition to the face values).
My Thoughts:
My intuition says that there should be a largest unrepresentable value. Empirical results seem to agree as well. I am just unsure of how to approach a proof. The traditional way of proving the coin problem (using some clever modular arithmetic to show that consecutive runs of arbitrary length occur in the sequence) won't work here though. Clearly this needs to be approached at an entirely different angle, but I don't know where to begin.
We claim that every integer $z \geq 49x^2y^2$ is in $S_{x,y}$. Since $\gcd(x,y)=1$, there exist $u,v\in\mathbb{Z}$ such that $ux-vy=1$. We may assume that $0< u\leq y$ and $0\leq v < x$. Now, note that the interval $I:=\left(\frac{zv}{x},\frac{zu}{y}\right)$ is of measure $$\frac{zu}{y}-\frac{zv}{x}=\frac{z}{xy}(ux-vy)=\frac{z}{xy}\geq 7\sqrt{z}>1+6\sqrt{z}\,.$$ Consequently, $I$ contains at least $\lceil6\sqrt{z}\rceil$ consecutive integers, whence $I$ contains an integer $p$ such that $\gcd(p,z)=1$, by the lemma below. Note that $\frac{zv}{x}<p<\frac{zu}{y}$.
To solve $n_1x+n_2y=z$ for $n_1,n_2\in\mathbb{N}$, we consider solutions $n_1=zu-p y>0$ and $n_2=px-zv>0$. We claim that $\gcd\left(n_1,n_2\right)=1$. First, from the assumption that $\gcd(p,z)=1$, there are $\mu,\nu\in\mathbb{Z}$ such that $p\mu+z\nu=1$. Now, take $a:=\mu v+\nu x$ and $b:=\mu u+\nu y$. We see that $$\begin{align} an_1+bn_2 &=(\mu v+\nu x)\big(zu-py\big)+(\mu u+\nu y)\big(px-zv\big) \\ &=z\big(u(\mu v+\nu x)-v(\mu u+\nu y)\big)+p\big(x(\mu u+\nu y)-y(\mu v+\nu x)\big) \\ &=\nu(ux-vy)z+\mu(ux-vy)p=\mu p +\nu z=1\,, \end{align}$$ which implies that $\gcd\left(n_1,n_2\right)=1$. Hence, $z\in S_{x,y}$.
Lemma: For every positive integer $N$, a set $L$ composed by consecutive integers such that $|L|\geq\lceil6\sqrt{N}\rceil$ must contain an integer $t$ which is coprime to $N$.
(The lemma can be generalized. For every $\epsilon>0$, there exists a constant $\kappa_\epsilon>0$ for which the following holds: for any $N \in \mathbb{N}$, $d\in\mathbb{N}$, and $r\in\mathbb{Z}$ such that $\gcd(N,d)=1$, a set $L$ composed by consecutive elements of the arithmetic progression $\big\{nd+r\big\}_{n\in\mathbb{Z}}$ with $|L| \geq \left\lceil \kappa_\epsilon N^\epsilon\right\rceil$ has an element $t\in L$ such that $\gcd(t,N)=1$. If $\epsilon \geq 1$, then trivially, $\kappa_\epsilon$ can be taken to be $1$. A proof for this generalized version with $0<\epsilon<1$ is done similarly to the one given below.)
Proof: Suppose that $N$ has (pairwise distinct) prime factors $p_1,p_2,\ldots,p_r \in\mathbb{N}$. Let $L$ be a set composed by $l$ consecutive integers, where $l\geq \lceil 6\sqrt{N}\rceil$.
Let $[r]:=\{1,2,\ldots,r\}$. For each subset $T$ of $[r]$, define $f_L(T)$ to be the number of elements of $L$ divisible by $p_j$ for all $j\in T$. Obviously, $\left\lfloor \frac{l}{\prod_{\lambda \in T} p_\lambda}\right\rfloor \leq f_L(T) \leq \left\lceil \frac{l}{\prod_{\lambda \in T} p_\lambda}\right\rceil$. Hence, the number of elements of $L$ which are coprime to $N$ is given by $$K_L:=\sum_{T\subseteq[r]}\,(-1)^{|T|}f_L(T)\,,$$ where we have used the Principle of Inclusion and Exclusion.
If $T\subseteq[r]$ is such that $|T|$ is even, $$f_L(T)> \frac{l}{\prod_{\lambda \in T} p_\lambda}-1=\frac{l}{\prod_{\lambda \in T} p_\lambda}-(-1)^{|T|}\,,$$ if $T\subseteq[r]$ is such that $|T|$ is odd, $$f_L(T)<\frac{l}{\prod_{\lambda \in T} p_\lambda}+1=\frac{l}{\prod_{\lambda \in T} p_\lambda}-(-1)^{|T|}\,.$$ This means $$ \begin{align} K_L &>\sum_{T\subseteq[r]}\,\left(\frac{(-1)^{|T|}l}{\prod_{\lambda \in T} p_\lambda}-1\right)=l\,\sum_{T\subseteq[r]}\frac{(-1)^{|T|}}{\prod_{\lambda\in T} p_\lambda}-\sum_{T\subseteq[r]}1 \\ &=l\,\prod_{j=1}^r\,\left(1-\frac{1}{p_j}\right)-2^r \geq 6\sqrt{N}\,\prod_{j=1}^r\,\left(1-\frac{1}{p_j}\right)-2^r \\ & \geq 6\sqrt{\prod_{j=1}^r\,p_j}\,\prod_{j=1}^r\,\left(1-\frac{1}{p_j}\right)-2^r=6\,\prod_{j=1}^r\,\left(\frac{p_j-1}{\sqrt{p_j}}\right)-2^r\,. \end{align}$$
If we order the positive primes as $q_1<q_2<q_3<\ldots$, then $$6\,\prod_{j=1}^r\,\left(\frac{p_j-1}{\sqrt{p_j}}\right) \geq 6\,\prod_{j=1}^r\,\left(\frac{q_j-1}{\sqrt{q_j}}\right)\,.$$ Observe that, for an integer $j \geq 4$, $\frac{q_j-1}{\sqrt{q_j}}>2$. By induction on $r$ (where the base cases $r=1,2,3$ must be manually checked), we conclude that $6\,\prod_{j=1}^r\,\left(\frac{q_j-1}{\sqrt{q_j}}\right)>2^r$ for every $r \in \mathbb{N}$. Thus, $$K_L > 6\,\prod_{j=1}^r\,\left(\frac{p_j-1}{\sqrt{p_j}}\right)-2^r \geq 6\,\prod_{j=1}^r\,\left(\frac{q_j-1}{\sqrt{q_j}}\right)-2^r > 0\,.$$ Ergo, $L$ contains an element $t$ such that $\gcd(t,N)=1$.
P.S. I believe that the largest integer $m(x,y)$ that is not in $S_{x,y}$ satisfies $m(x,y) \in \Theta(xy)$. What I know is that $m(x,y)\in\Omega(xy)$ and $m(x,y) \in O\left((xy)^{1+\varepsilon}\right)$ for all $\varepsilon>0$. However, it seems like an extremely difficult problem to find an explicit or an asymptotic formula for $m(x,y)$.
EDIT: After some experiments, it seems like $m(x,y)\in\Theta\big((xy)\,\ln(xy)\big)$. A plot to illustrate this conjecture is given below. Here, $x,y\in\{1,2,\ldots,50\}$. The scattered plot in blue is the actual $xy$ versus $m(x,y)$. The red curve is the estimate $m(x,y)=(xy)\,\ln(xy)$. If you have an efficient way to write a code to determine the relationship between $xy$ and $m(x,y)$, please share it with us. My code takes quite long to run.