Why isn't it obviously

It is nearly yes ? as you need to use linear combinations rather then sums. If $y=\sum_i e_i \otimes b_i$ then $x=yy^* =\sum_{i,j} e_ie_j^* \otimes b_ib_j^*$. This leaves you with two types of summands whose positivity is clear:

(1) $e_ie_i^*\otimes b_ib_i^*$

(2) $e_ie_j^* \otimes b_ib_j^* + e_je_i^* \otimes b_jb_i^*$ whose positivity is clear by the elementary calculation that boils down to multilinearization of $(\alpha e_i+\beta e_j)(\alpha e_i+\beta e_j)^* \otimes (\gamma b_i+\delta b_j)(\gamma b_i+\delta b_j)^*$. (Just write the second summand as a linear combination of these guys.)

Clearly, you need to use subtractions to clear up the summands of type (2).

Why isn't it obviously yesif $x$ is positive? Pick an algebraic basis of $A$ such that if $e$ is an element of the basis then so is $e^*$. Writing in this basis, If $x=\sum_i y=\sum_i e_i \otimes b_i = x^* b_i$ then $x=yy^* =\sum_i e_i^\sum_{i,j} e_ie_j^* \otimes b_i^*$b_ib_j^*$. This leaves you with two casestypes of summands whose positivity is clear:

(1) $e_i^*=e_i$ and then $b_i^*=b_i$.e_ie_i^*\otimes b_ib_i^*$

(2) $e_i^*= e_j \neq e_i$ and then $b_i^*=b_j$. It remains to write $e_i e_ie_j^* \otimes b_i b_ib_j^* + e_j e_je_i^* \otimes b_j = ((e_i+e_j) b_jb_i^*$ whose positivity is clear by the elementary calculation that boils down to multilinearization of $(\alpha e_i+\beta e_j)(\alpha e_i+\beta e_j)^* \otimes (b_i+b_j) - I(e_i-e_j)\gamma b_i+\delta b_j)(\gamma b_i+\delta b_j)^*$. (Just write the second summand as a linear combination of these guys.)\otimes I(b_i-b_j))/2$ where $I=\sqrt{-1}$.

Why isn't it obviously yes if $x$ is positive? Pick an algebraic basis of $A$ such that if $e$ is an element of the basis then so is $e^*$. Writing in this basis, $x=\sum_i e_i \otimes b_i = x^* =\sum_i e_i^* \otimes b_i^*$. This leaves you with two cases:

(1) $e_i^*=e_i$ and then $b_i^*=b_i$.

(2) $e_i^*= e_j \neq e_i$ and then $b_i^*=b_j$. It remains to write $e_i \otimes b_i + e_j \otimes b_j = ((e_i+e_j) \otimes (b_i+b_j) - I(e_i-e_j)) \otimes I(b_i-b_j))/2$ where $I=\sqrt{-1}$.