Does Smith normal form imply PID?

The implication is false without the assumption that R is Noetherian, because finite matrices don't detect enough information about infinitely generated ideals.

For example, let R be the ring $$ \bigcup_{n \geq 0} k[[t^{1/n}]] $$ where $k$ is a field (an indiscrete valuation ring). Any finite matrix with coefficients in R comes from a subring $k[[t^{1/N}]]$ for some large $N$, and hence can be reduced to Smith normal form within this smaller PID.

However, the ideal $\cup (t^{1/N})$ is not principal.

If every matrix has a Smith normal form, then every finitely generated $R$-submodule $M$ of $R^n$ satisfies $R^n/M$ is a finite direct sum of modules isomorphic to $R/aR$. If $R$ is Noetherian this implies that every finitely generated module is a direct sum of modules of the form $R/aR$. So if $I$ is a maximal ideal of the Noetherian $R$ then $R/I$ is a simple module, so if $R/I\cong R/aR$ then $I=aR$ is principal. So in a Noetherian ring with Smith normal form for all matrices, every maximal ideal is principal. Does this imply that all ideals are principal?....I'm not sure :-)

Work on ring-theoretic generalizations of Hermite/Smith normal forms goes way back, but made it into the mainstream via classic papers by Helmer and Kaplansky. Nowadays such rings are called elementary divisor rings, or rings with elementary divisors (r.e.d.) or Helmer rings, etc. A search on such terms, and for citations of Kap's classic paper [1] should quickly answer all your questions and then some.

[1] I. Kaplansky, "Elementary divisors and modules," Trans. Am. Math. Soc., 66, 464-491. (1949).
http://www.ams.org/journals/tran/1949-066-02/S0002-9947-1949-0031470-3/S0002-9947-1949-0031470-3.pdf