I think this is easy to see, using Gelfand Transform. Using the transform we can see $spec(a)=\{\phi(a)| \phi\in Spec(A)\}$ thus $spec(a+b)=\{\phi(a)+\phi(b)|\phi\in Spec(A)\}$ Now if both $a$ and $b$ are positive then clearly $\phi(a)$ and $\phi(b)$ sperately are non-negative and thus their sum is also.
I think this is a nice proof but I was wondering if there is a way of proving this without appealing to the Gelfand transform.
The following characterization of positivity will do the trick.
Let $A$ be unital (otherwise consider the minimal unitization $\tilde{A}$ instead of $A$), let $x \in A$ be self-adjoint and $\lambda \geq \|x\|$. Then $x$ is positive if and only if $\|\lambda 1_A -x\| \leq \lambda$.
Proof. Consider the constant function $1$ and the identity function $id$ in $C(spec(x))$. Since $x$ is self-adjoint and $\lambda \geq \|x\|$ we have $spec(x) \subseteq \left[-\lambda,\lambda\right]$; moreover observe that $spec(x) \geq 0$ if and only if $\|\lambda 1 - id\|_{\infty} \leq \lambda$. Then by functional calculus we infer that $x$ is positive if and only if $\|\lambda 1_A -x\| = \|\lambda 1 - id\|_{\infty}\leq \lambda$. QED
Now let's go back to the core discussion:
Let $x,y$ be positive elements of $A$, then $x+y$ is positive.
Proof. By the previous characterization we have in particular that $\| \|x\| -x \| \leq \|x\|$ and the same for $y$. Now $ \| \|x\| + \|y\| -(x+y) \| \leq \| \|x\| -x \| + \| \|y\| -y \| \leq \|x\| + \|y\|$. Again by the characterization above (with $\lambda = \|x\|+\|y\|$, where the triangle inequality guarantees that $\|x+y\| \leq \lambda$) we conclude that $x+y$ is positive. QED