This answer deals with the case that $\phi$ is non-unital. In this case, $\phi$
must be of the form $\phi(a)=ha$, where $h$ is a positive element in the center
of $A$.

Unfortunately, the solution I've got is somewhat long (hopefully correct): Let us assume that $\phi$ is contractive (otherwise one can rescale). Since $\phi(a)\leq a$ for any positive element $a$, $\phi$ preserves orthogonality of positive elements. There's a nice structural result for such maps proven in ``Completely positive maps of order zero", by Winter and zacharias. By Theorem 2.3 of that paper $\phi$ has the form $\phi(a)=h\pi(a)$, where $h=\phi(1)$,
$$\pi:A\to M(C^*(\phi(A)))$$

is a unital homomorphism into the multiplier algebra of $C^*(\phi(A))$ and $h$ commutes with the image of $\pi$.

From $\phi(a)\leq a$ one gets $h\pi(a)\leq a$ for any positive $a$.
Suppose that $\pi(b)=0$, with $b\in A_+$ positive contraction. Then $h=h\pi(1-b)\leq 1-b$.
Similarly, $h\leq 1-b^{1/n}$ for all $n$. This implies that $hb=0$. So $h$ is orthogonal to $\ker \pi$. In particular, $\pi$ is injective on $\overline{hAh}$.

Let us show that $\pi$ is the identity on $\overline{hAh}$. Taking $n$-root in $h\pi(h^n)\leq h^n$ we get $h^{1/n}\pi(h)\leq h$. Since $h$ is a strictly positive element of $C^*(\phi(A))$ and $\pi(h)$ is a multiplier for that algebra, we have that $h^{1/n}\pi(h)\to \pi(h)$ in the strict topology. So, $\pi(h)\leq h$, and so $\pi$ maps $\overline{hAh}$ into itself. Passing to the limit in $h^{1/n}\pi(a)\leq a$, with $a\in \overline{hAh}$, one gets $\pi(a)\leq a$ for all such $a$. But if $\alpha$ and $\beta$ are homomorphisms such that $\alpha\leq \beta$ on all positive elements, then $\alpha+\tilde\alpha=\beta$, where $\tilde\alpha$ is a homomorphism with orthogonal range to $\alpha$. (There's probably a reference for this; I'll skip the argument to keep this answer short.) In our case, since $\pi$ is injective, we get $\pi(a)=a$ for $a\in \overline{hAh}$. In particular, $\pi(h)=h$.

We have that
$$
\phi(a)=h\pi(a)=\pi(h^{1/2} a h^{1/2})=h^{1/2} a h^{1/2}
$$for all $a\in A$. Finally, let us show that $h$ is in the center of $A$.
We have $\pi(ah-ha)=\pi(a)\pi(h)-\pi(h)\pi(a)=0$. So
$ha-ah\in \ker \pi$. It was argued above that this implies
$h(ha-ah)=0$. So $h^2a=ah^2$. So $h$ is in the center.