Let us define a random variable (RV) Z = X + Y where X and Y are two independent RVs. If f_X(x) and f_Y(y) are probability density functions of X and Y, then what can we say about f_Z(z), the pdf of Z? A rigorous double “E” graduate course in stochastic processes is usually sufficient to answer this question. It turns out f_Z(z) is the convolution of the densities f_X(x) and f_Y(y). See this (p.291) for more.

It is tempting to ask the counter question: if f_Z(z) = f_X(x)*f_Y(y), then does it imply Z = X + Y where X and Y are independent RVs? I found the answer in this amazing text:
Counterexamples in Probability by Jordan Stoyanov. The book is an adorable compilation of some 300 counterexamples to the probability questions which might be bothering you during a good night-sleep.

So, the answer to my question is no. The counterexample is using the Cauchy distributions. It turns out that the convolution of two Cauchy distributions is always a Cauchy distribution whether X and Y are independent or not.

Coming to think of the convolution of pdfs, our favorite website has a list of convolution of common pdfs.