Just so that the question leaves the unanswered list: Yes, commutative $C^*$-algebras are stably finite and here is a short argument:
It is a known fact that, if $A$ is a unital $C^*$-algebra, then $A$ is finite if and only if $s^*s=1_A\implies ss^*=1_A$ for all $s\in A$, i.e. every isometry is a unitary.
Let $n\geq1$. We have that $M_n(C(X))\cong C(X,M_n(\mathbb{C}))$. Now let $s\in C(X,M_n(\mathbb{C}))$ be an isometry, i.e. $s^*s=1_{C(X,M_n)}$, i.e. $s(x)^*s(x)=1_{M_n}$ for all $x\in X$. Since $M_n(\mathbb{C})$ is finite (obviously) we conclude that $s(x)s(x)^*=1_{M_n}$ for all $x\in X$, thus $ss^*=1_{C(X,M_n)}$ and this shows that $C(X)$ is stably finite.
A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.
Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.
Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.
Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.
Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.
Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.
Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.
Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.
Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.
Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.
Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.
Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.
Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.
Best Answer
Consider $$ A = \{f\in C([0, 1], \mathcal T): f(0)\in \mathbb C\cdot 1\}, $$ where $\mathcal T$ is the Toeplitz algebra. Every isometry $f$ in $A$ gives rise to a path of isometries in $\mathcal T$ beginning with a scalar. Observing that all isometries in the Toeplitz algebra have finite Fredholm index, and also that the index is continuous, we can easily show that $A$ is finite. However the map $$ f\in A\mapsto f(1)\in \mathcal T $$ is a surjection from $A$ to a non-finite algebra.
EDIT. Here is an argument to show the finiteness of $A$ which does not rely on the theory of Fredholm operators:
Lemma. Let $B$ be a unital C*-algebra. Denote by $\mathscr I(B)$ the set of all isometries in $B$, and by $\mathscr U(B)$ the subset formed by the unitary elements. Then $\mathscr U(B)$ is clopen in $\mathscr I(B)$.
Proof. It is well known that the set $\text{GL}(B)$ formed by all invertible elements is open. Since $$ \mathscr U(B) = \mathscr I(B) \cap \text{GL}(B), $$ it follows that $\mathscr U(B)$ is open in $\mathscr I(B)$. To show that $\mathscr U(B)$ is also closed, suppose by contradiction that the sequence $\{u_n\}_n$ of unitaries converges to a proper isometry $s$ (meaning that $ss^*\neq 1$). The element $$ p = 1-ss^* $$ is then a non-trivial projection and, clearly, $ps=0$. We then deduce that $$ 0 = ps = \lim_{n\to \infty }pu_n, $$ but this is a contradiction since $\|pu_n\|=\|p\| = 1$. QED
This said, suppose that $f$ is an isometry in $A$, so we have that $f(t)$ lies in $\mathscr I(\mathcal T)$, for all $t$ in $[0, 1]$.
Since $f(0)$ is necessary a scalar, we have that $f(0)\in \mathscr U(\mathcal T)$, but then $f(t)\in \mathscr U(\mathcal T)$, for every $t$ because $[0,1]$ is connected.
This shows that $f$ is unitary and hence that $A$ is finite.