Can the statement be prove using the First Isomorphism Theorem?
(1) Take $A,B$ rings and $f:A\rightarrow B$ a morphism.
(2) Define $g:M_n(A)\rightarrow M_n(B)$ having $A=(a_{ij}), B=(b_{ij})$ and $g(a_{ij})=b_{ij}$. Then $g$ is also a ring morphism.
(3) By the first isomorphism theorem $\frac{M_n(A)}{\ker{g}}\simeq Im(M_n(A))= M_n(B).$
(4) $\frac{A}{\ker{f}}\simeq Im(A)=B$
(5) $\frac{M_n(A)}{\ker{g}}\simeq M_n \left(\frac{A}{\ker{f}}\right)$.
I know $\ker{f}$ and $\ker{g}$ are ideals of $A$ and $M_n(A)$ but the theorems is stated for ideals in general. I think the theorem could be proved if it is possible to show that given a ring $A$ and an ideal $I$ then there is always morphism $\phi$ such that $\ker \phi$=I.
As you suggest the first isomorphism theorem does indeed give the result. Take $f\colon A \to A/I$ to be the natural quotient homomorphism. This gives a surjective homomorphism $g\colon\mathbb M_n(A) \to \mathbb M_n(A/I)$ whose kernel is $\mathbb M_n(I)$.