Does homotopy equivalence to a wedge of spheres imply shellability?
No. There are unshellable n-spheres for all n≥3:
Lickorish, W. B. R., Unshellable triangulations of spheres, Eur. J. Comb. 12, No. 6, 527-530 (1991). ZBL0746.57007.
As @j.c. says, the answer is "no", and it is good to know that there are non-shellable triangulations of spheres.
However, it is even easier to find counterexamples, that are not shellable because of topological obstructions. I'll give two examples:
The annulus is homotopy equivalent to $S^1$, but no triangulation is shellable (since a shellable 2-dimensional complex is topologically a bouquet of 2-spheres, not 1-spheres).
More deeply, the wedge of 2 or more spheres of dimension at least 2 is never shellable, nor even Cohen-Macaulay. (Note the lack of the word "homotopy type" in the previous sentence!) This is because if you take two $d$-spheres and glue them together at the point $v$, then the link of $v$ is disconnected of dimension $d-1$, hence not shellable. Since the link of every vertex in a shellable complex is shellable, this gives the desired non-shellability.
The fragility of shellability in this respect can be seen as a good thing: there are lots of tools to compute homotopy type of a complex, but shellability (when carefully examined) tells you a fair bit about the "homeomorphism type" of the complex.
By the way, the smallest dimension of a non-shellable sphere is 3. The 1978 papers of Danaraj and Klee are one reference for this; the first paper I'm aware of that uses the word shelling, "Isotopy in 3-manifolds- Isotopic deformations of 2-cells and 3-cells" by Donald Sanderson, is another.