Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $\mathcal{S}=\{ax+by\mid x,y\ge0 \text{ and } x,y\in \mathbb{Z}\}$. 
Let $a,b\in \mathbb{N}$ where $\gcd(a,b)=1$. Describe the set $$\mathcal{S}=\{ax+by \mid  x,y\geq0 \hspace{0.25cm} \text{ and } \hspace{0.25cm} x,y\in \mathbb{Z}\}.$$

Since $a,b\in \mathbb{N}$ and $\gcd(a,b)=1$, then there exists $x',y'\in\mathbb{Z}$ such that $$ax'+by'=1.$$ Let $\alpha\in\mathcal{S}$. Then  $$\alpha=ax+by$$ where $a,b\in \mathbb{N}$ with $\gcd(a,b)=1$ and $x,y\in \mathbb{Z}$ with $x,y\geq 0$.
If we were to multiply the first equation by $\alpha$ we will get $$\alpha ax'+\alpha by'=\alpha.$$ But $\alpha=ax+by$. So, $$\alpha ax'+\alpha by'=\alpha=ax+by.$$ We can rewrite this equation as follows:
$$a(\alpha x'-x)=b(y-\alpha y').$$ Than $a(\alpha x'-x)| b$ or $a(\alpha x'-x)| (y-\alpha y')$ or $a(\alpha x'-x)| b(y-\alpha y')$. 
From here I am unsure how to proceed. Any tips

I'm not certain about your divisibility conditions. It seems like only the third will hold. I proceed slightly differently.
Let's assume, without loss of generality, that $1<a<b$. Then we can construct an array of non-negative integers the is $a$ units wide and $b$ units long. This gives us a way to count all numbers from $1$ to $ab$. It is clear that all positive multiples of $a$ will be in $S$. 
Ask yourself this: what happens to multiples of $b$? For $1\leq k\leq a-1$ we have $bk$ will land in some nonzero congruence $\mod a$. In fact, each will be distinct (by coprimality). 
In addition, every element beneath $bk$ will be in $S$, since you just add multiples of $a$ to get to the next row. Once you get to $(a-1)b$, you have all congruence class filled $\mod a$. If you subtract $a$ from this value ($ab-b-a$), you have the last element that is not in $S$. 
Every natural number greater than or equal to $ab-b-a+1$ will be in $S$! Let's see this with $a=3$ and $b=5$
\begin{align*}
1\hspace{1cm}2\hspace{1cm}s_1\\
4\hspace{.95cm}s_2\hspace{.9cm}s_3\\
7\hspace{.95cm}s_4\hspace{.9cm}s_5\\
s_6\hspace{.85cm}s_7\hspace{.9cm}s_8\\
s_9\hspace{.8cm}s_{10}\hspace{.7cm}s_{11}
\end{align*}
where the $s$ values are 3, 5, 6, 8, 9, 10, 11, 12, 13, 14, and 15. Once we get to $3\cdot5-3-5+1=8$, we intersect all integers.
