Positivity in non-unital $C^*$-algebras For unital algebras, we say that an element $b$ is positive if there exists an element $a$ such that $b=a^*a$ if and only if $\sigma(b)\subseteq \mathbb{R}_+$. In the non-unital case, how would one define positivity? Also, how would this fit together with the notion above when passing to unitizations?
Secondly, when introducing complete positivity of maps, the framework is usually given as an operator system (i.e., closed, selfadjoint subspaces of unital algebras). Could we do without this extra assumption and meaningfully introduce this concept for arbitrary subspaces?
 A: If $A$ is non-unital, and $a \in A$, we define
$$\sigma_A(a):= \sigma_{\widetilde{A}}(a)$$ where $\widetilde{A}$ is the unitisation of $A$.
We then define:

An element $x \in A$ is called positive if it is self-adjoint and if $\sigma_A(x)\subseteq [0, \infty[$.

Note the self-adjointness assumption in the above definition (which you left out in your characterisation of positive element!).
A natural question is to see what happens if $A$ is already unital. Then for $a \in A$, we have
$$\sigma_{\widetilde{A}}(a) = \{0\}\cup \sigma_A(a).$$
Thus it does not matter if you examine positivity in $A$ or in $\widetilde{A}$ (regardless if $A$ is unital), as both notions coincide.
And yes, the notion of (complete) positive map makes sense for arbitrary subspaces of (not necessarily unital) $C^*$-algebras. Indeed, let $M\subseteq A$ a subspace of a $C^*$-algebra and $B$ a $C^*$-algebra. Then we call $f: M \to B$ positive if $f(m)$ is positive in $B$ whenever $m$ is positive in $M$ (meaning that it is positive in $A$).

In reply to a comment:

Assume $A$ is non-unital and $a \in A$ is positive. Then there exists a positive element $b \in A$ with $a=b^2$.

Proof: Functional calculus gives us a $*$-isomorphism
$$C(\sigma_{\widetilde{A}}(a)) \to C^*(1_{\widetilde{A}}, a)\subseteq \widetilde{A}: f \mapsto f(a).$$
Apply this to the function $f= \sqrt{-}$ and let $b:= f(a)$. But $f(0)=0$, so a routine argument implies that in fact $b \in C^*(a)$ (hint: show that $f$ can be uniformly approximated by polynomials with constant term $0$). Since $C^*(a)\subseteq A$, you are done. $\quad \square$
