No, e.g. in $\rm\:\Bbb Q[x,y],\:$ $\rm\:gcd(x,y) = 1\:$ but $\rm\: x r + y s = 1\:\Rightarrow\: 0 = 1\:$ by evaluating at $\rm\,x,y=0.$
Here are $\,2\,$ proofs, using $(1)$ existence and uniqueness of prime factorizations, and $(2)$ gcds.
$(1)$ $\rm\ a\mid bc\:$ so $\rm\:ad = bc\:$ for some $\rm\:d.\:$ By existence, we can factor all four terms into prime factors. By uniqueness, the same multiset of primes occurs on both sides (up to unit factors / associateness). So all of the primes in the factorization of $\rm\:a\:$ must also occur on the RHS, necessarily in the factorization of $\rm\:c,\:$ since, by hypothesis, $\rm\:a,b\:$ are coprime, hence share no prime factors. Therefore $\rm\:a\mid c\:$ since all of its prime factors (counting multiplicity) occur in $\rm\:c.\:$ Note that this inference can be expressed purely multiset-theoretically: $\rm\: A\cap B = \{\,\},\ \ A \cup D\,=\, B\cup C\:$ $\:\Rightarrow\:$ $\rm\:A\subset C,\:$ where $\rm\:C =$ multiset of primes in the unique prime factorization of $\rm\:c,\:$ and similarly for $\rm\:A,B,D.$
$(2)\ $ Using gcds: $\rm\ \ a\mid ac,bc\:\Rightarrow\:a\mid \color{#0A0}{(ac,bc)=(a,b)c} = c\:$ by $\rm\:(a,b)=1,\:$ and by the following
Theorem $\rm\ \ \ (a,b)\ =\ (ac,bc)/c\quad \rm\color{#0A0}{(\,gcd\ distributive\ law)}$
Proof $\rm\quad\ d\mid a,b\ \iff\ dc\mid ac,bc\ \iff\ dc\mid (ac,bc)\ \iff\ d\mid(ac,bc)/c$
The theorem holds more generally in any domain as long as $\rm\ (ac,bc)\ $ exists. See my post here for further discussion of this property and its relationship with Euclid's Lemma.