If $A$ and $B$ have non-negative eigen values then can we conclude that $A+B$ has non-negative eigen values?

The eigenvalues can be negative or non-real.

Take $$A=\begin{bmatrix}1&a\\b&1\end{bmatrix},B=\begin{bmatrix}1&c\\d&1\end{bmatrix}.$$ Then $A$ and $B$ have positive real eigenvalues as long as $0<ab<1$ and $0<cd<1$, but the eigenvalues of $A+B$ are negative if $$(a+c)(b+d)>4$$ and non-real if $(a+c)(b+d)<0$.

For the first case to hold, take $a=d=3$ and $b=c=1/4$. For the second, take $$(a,b,c,d)=\left(3,\frac14,-\frac14,-3\right).$$

Take $A=\begin{pmatrix} 1& 2\\0&1\end{pmatrix}$, $B=\begin{pmatrix} 1& 0\\2&1\end{pmatrix}$. Then both have $1$ as unique eigenvalue but $A+B$ has zero as eigen value.

Take $C=\begin{pmatrix} 1& 1\\0&1\end{pmatrix}$, $D=\begin{pmatrix} 1& 0\\-1&1\end{pmatrix}$. Then both have $1$ as unique eigen value but $C+D$ has no real eigenvalue.

Take $E=\begin{pmatrix} 1& 1\\0&2\end{pmatrix}$, $F=\begin{pmatrix} 2& 0\\-1&1\end{pmatrix}$. Then both have $1,2$ as distinct eigen value but $E+F$ has no real eigenvalue.

Take $G=\begin{pmatrix} 1& 5\\0&2\end{pmatrix}$, $H=\begin{pmatrix} 2& 0\\5&1\end{pmatrix}$. Then both have $1,2$ as distinct eigen value but $G+H$ has $8,-2$ as eigenvalue.