As @julien points out, the function $\frac{\sin(x+y)}{x+y}$ is not defined in a neighbourhood of the origin, so the limit does not exist. But there is a natural extension of the function which is defined on a neighbourhood of the origin, and for which the limit exists and equals 1. (There are lots of extensions, but this one seems most natural to me.) Define $f:\mathbb{R}^2\to\mathbb{R}$ by $$f(x,y)=\left\{
\begin{array}{cl}
\frac{\sin(x+y)}{x+y} & x+y\neq0; \\
1 & x+y=0.\end{array}\right.$$
Then $\lim_{(x,y)\to(0,0)}f(x,y)=1$. An $\epsilon-\delta$ proof relies on the one variable result $$\lim_{u\to 0}\frac{\sin u}{u}=1.$$
That is, given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon.$$
It is helpful to notice that this can be rephrased as follows: given $\epsilon>0$, there exists $\delta>0$ such that $$0<|u|<2\delta \Rightarrow \left|\frac{\sin u}{u}-1\right|<\epsilon. \qquad (*)$$
We want to prove that given $\epsilon>0$, there exists a positive $\delta$ such that $$0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-1|<\epsilon.$$
So let $\epsilon>0$. Choose $\delta$ so that $(*)$ holds true, and let $0<\sqrt{x^2+y^2}<\delta$. Then $$|x+y|\leq|x|+|y|\leq\sqrt{|x|^2+|y|^2}+\sqrt{|x|^2+|y|^2}=2\sqrt{x^2+y^2}<2\delta.$$ Then by (*) (for $x+y\neq0$) and by the definition of $f$ (for $x+y=0$), we have $|f(x,y)-1|<\epsilon$ as required.
The same argument works if the function is defined as $g:\mathbb{R}^2\setminus\{(x,y):x+y=0\}\to\mathbb{R}$ with $g(x,y)=\frac{\sin(x+y)}{x+y}$. With a slight amendment, the same result also holds for $$f(x,y)=\left\{
\begin{array}{cl}
\frac{\sin(x+y)}{x+y} & x+y\neq0; \\
h(x) & x+y=0,\end{array}\right.$$
where $h$ is any function defined in a neighbourhood of $x=0$ which has $\lim_{x\to0}h(x)=1$.
