If coprime elements generate coprime ideals, does it imply for any $a,b\in R$ that $\langle a\rangle+\langle b\rangle=\langle \gcd (a,b)\rangle$?

Yes, $(2)\Rightarrow (1)$ holds. If $\rm\:d = gcd(a,b)\:$ then $\rm\:(a,b)\:=\: d\:\!(a/d,b/d) = d\!\:(1) = (d)\:$ by $(2)$. Hence a GCD domain is Bezout iff coprime elements are comaximal (a domain is called Bezout if two-generated (so finitely generated) ideals are principal).

Remark $\ $ For a comprehensive survey of integral domains closely related to GCD domains see D.D. Anderson: GCD domains, Gauss' lemma, and contents of polynomials, 2000. See also

Theorem $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$

$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le 1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1, $ i.e. $ $ coprime $\Rightarrow$ comaximal
$(5)\ \ $ $\rm D$ is Bezout
$(6)\ \ $ $\rm D$ is a $\rm PID$

Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\ $ where $\rm\,p_i,\,P\,$ denote primes $\neq 0$

$(1\Rightarrow 2)$ $\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$
$(2\Rightarrow 3)^{\phantom{|^i}}\!\!\!$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)^{\phantom{|^i}}\!\!\!$ $\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)^{\phantom{|^|}}\!\!\!$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)^{\phantom{|^|}}\!\!\!$ $\ $ Ideals $\neq 0$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)^{\phantom{|^|}}\!\!\!$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max

A domain in which every sum of two (or finitely many) principal ideals is again a principal ideal is called a Bézout domain (curiously the English write an accent in that name, whereas the French, aware of the fact that writing accents was very haphazard at the time, don't). A Bézout domain is always a GCD domain, but the converse is not true. It is easy to see that the generator of $\langle a\rangle+\langle b\rangle$ is necessarily a gcd of $a$ and $b$. Your question is therefore whether a GCD domain in which (2) holds is necessarily a Bézout domain. The answer is yes, as indicated in the comment by Joel Cohen: when $d=\gcd(a,b)$, the elements $a/d$ and $b/d$ cannot have a common factor, so $1$ is a gcd of $a/d$ and $b/d$, and by (2) there exist $s,t$ with $s(a/d)+t(b/d)=1$, which implies $as+bt=d$ and therefore (1).