In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult:

- A construction of Kaplansky (Rings of operators, Thm 26) shows that if $\mathcal{A}$ is a $*$-algebra with the property that for every $A \in M_n(\mathcal{A})$ the matrix $1 + A^*A$ is invertible, then every idempotent in $M_n(\mathcal{A})$ is equivalent to a projection. A $C^*$-algebra and in particular $C(M)$ clearly has this feature by spectral calculus for every $n \ge 1$.

Suppose now that $\mathcal{A}$ is such a $^*$-algebra satisfying this condition for all $n$ (†).

On every finitely generated projective module $\mathcal{E}$ over $\mathcal{A}$, written without restriction as $\mathcal{E} = P \mathcal{A}^n$ with $P = P^2 = P^* \in M_n(\mathcal{A})$, the restriction $\langle . , .\rangle$ of the canonical Hermitian algebra-value inner product of the free module $\mathcal{A}^n$ is still positive (in the strongest possible sense) and non-degenerate (in the strongest possible sense that the musical homomorphism is in fact an anti-isomorphism to the dual module). This gives immediately the existence of positive algebra-valued inner products on finitely generated projective modules over such algebras.

Now suppose the algebra satisfies an additional feature: suppose $H \in M_n(\mathcal{A})$ is an invertible positive element. Then $H$ has a positive square root $H = \sqrt{H}^2$ with the property that $\sqrt{H}$ commutes with all matrices which commute with $H$ (‡). Again, a $C^*$-algebra clearly satisfies this for all $n$ by spectral calculus.

Now suppose that on the fgpm $\mathcal{E} = P \mathcal{A}^n$ you have another positive algebra-valued inner product, denoted by $h$. On the complement $P^\bot = (1 - P)\mathcal{A}^n$ we use the restriction of the canonical algebra-valued inner product to obtain a new inner product, still denoted by $h$, on the direct sum. This is still positive and has all required (very strong) non-degeneracy properties.

- By matrix calculus with free modules one obtains then a matrix $H \in M_n(\mathcal{A})$ with
\begin{equation}
h(x, y) = \langle x, Hy\rangle
\end{equation}
for all $x, y \in \mathcal{A}^n$. Since by assumption $H = \sqrt{H}^2$ the square root $U = \sqrt{H}$ provides an isometry between the two inner products on the free module. Since by assumption the square root commutes with everything commuting with $H$ and since by construction $H$ commutes with $P$ (check this!) $U$ restrict to an isometry between $h$ and $\langle ., .\rangle$ on the original $\mathcal{E}$.

In conclusion: for $^*$-algebras (like e.g. $C^*$-algebras) satisfying the above two properties (†) and (‡), every finitely generated projective module carries a unique-up-to-isometry positive algebra-valued inner product.

As a side remark: many other types of $^*$-algebras satisfy these properties as well like e.g. the smooth functions on a manifold etc. So the same proof also implies (via Serre-Swan in the smooth context) that on smooth vector bundles you always have a unique-up-to-isometry positive fiber metric.

Now, to answer you question: the uniqueness gives you the desired result that every algebra-valued inner product comes from a positive fiber metric. Indeed, a positive fiber metric meets the above properties and the isometry $U$ maps fiber metrics to fiber metrics as it is algebra-linear.