Let $A$ be a finite, non-unital C*-algebra, $s\in A$ a strictly positive element and $a\in A$ a positive element that is Cuntz-equivalent to $s$, i.e. there exist sequences $\{x_n\},\{y_n\}\subset A$ such that $\Vert x_nax_n^*-s\Vert\to0$ and $\Vert y_nsy_n^*-a\Vert\to 0$. Is $a$ strictly positive?
In the unital case (even in the infinite case), a positive element $s$ is strictly positive if and only if it is invertible, and therefore it is Cuntz-equivalent to the unit of the algebra. Now if $a$ is Cuntz-equivalent to $s$, then it is also Cuntz equivalent to 1, which means that there exists a sequence $\{x_n\}$ in the unital algebra such that $$\Vert x_nax_n^*-1\Vert\to0.$$ In the finite case this implies that the sequence $\{x_n\}$ is eventually left-invertible (hence invertible because of the finiteness of the algebra). In turn $a$ is also left-invertible, hence invertible, and therefore strictly positive.
For the non-unital case I was thinking of considering the support projection $p_s$ of $s$ and conclude that $p_s-\overline p_a\neq 0$ if $a$ is not strictly positive, whence there exists $b$ with $p_b = p_s-\overline p_a$ such that $a+b\precsim a$ and $ab=0$, which is impossible in a finite C*-algebra. But this way of thinking seems overly complicated to me.
