If $A$ has stable rank one then this can be deduced from the following result:
If $B$ has stable rank one and $a,b\in B_+$ are Cuntz equivalent then
the right ideals $\overline{aB}$ and $\overline{bB}$ are isomorphic as Hilbert modules. This is Theorem 3 of

Coward, Kristofer T.; Elliott, George A.; Ivanescu, Cristian. The Cuntz semigroup as an invariant for $C^*$-algebras. J. Reine Angew. Math. 623 (2008), 161--193.

See also Proposition 1 of

Ciuperca, Alin; Elliott, George A.; Santiago, Luis. On inductive limits of type-I $C^*$-algebras with one-dimensional spectrum. Int. Math. Res. Not. IMRN 2011, no. 11, 2577--2615.

It's a little awkward to translate this into an answer to the question. Let me try.

First, this: Let $p,q\in M(B)$ be multiplier projections such that $pB\cong qB$ as Hilbert modules. Then $p$ is Murray-von Neumann equivalent to $q$. Proof: Let $v\colon pB\to qB$ be an isomorphism. Extend $v$ to $B$ setting it to 0 on $(1-p)B$. Then $v\in M(B)$ (regarded as the adjointable operators on $B$) and $v^*v=p$, $vv^*=q$.

Now let $B$ be separable and of stable rank one. Let $\phi$ be an approximately inner automorphism extended to $M(B)$. Let us show that any projection $p\in M(B)$ is Murray-von Neumann equivalent to $\phi(p)$. It suffices to show that $pB\cong \phi(p)B$. Let $c\in B$ be a strictly
positive element of $pBp$ (exists since $B$ is separable). Then $pB=\overline{cB}$.
Since $c^{1/n}\uparrow p$ strictly and $\phi$ is strictly continuous, we also have $\phi(p)B=\overline{\phi(c)B}$. But $c$ and $\phi(c)$ are approximately unitarily equivalent, so they are Cuntz equivalent. We can use the theorem recalled above.

To answer the question, we take $B=A\otimes \mathcal K$ and $p=1\otimes e_{11}$.
Notice that to answer the question we only need $A$ $\sigma$-unital rather than separable.

Remark: There exist examples of positive elements $a,b$ in an $A$ (of sr=2) that are approximately unitarily equivalent and yet $\overline{aA}$ and $\overline{bA}$ are not isomorphic as Hilbert modules. However, (1) $A$ is non-simple and they way the examples work they can't be tweaked to get this, (2) The unitary conjugations moving $a$ to $b$ don't seem to converge pointwise to an automorphism. So there is a chance that the question has a positive answer without sr1.