= Solution
Use finitely generated right <projective modules> throughout. The <group> <K0 of a ring> is the <Grothendieck group> of their direct-sum <monoid>, with $[P\oplus Q]=[P]+[Q]$. The <group> <K1 of a ring> is $\mathrm{GL}_\infty(A)/E_\infty(A)$, equivalently the <abelianization> of the <stable general linear group>. Stabilization is $g\mapsto\operatorname{diag}(g,1)$; $E_\infty$ is generated by elementary <transvections>. The <stable elementary subgroup> is the <commutator subgroup>: elementary <transvections> are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.
Write $A_0$ for the lower-right <ring>, printed as $A'$ in the source. A <Milnor square> has unital commuting maps, is a <pullback>
$$
A=\{(a_1,a_2)\in A_1\times A_2:j_1(a_1)=j_2(a_2)\},
$$
and has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.
We first establish the patching fact used in the proof. For projective $P_i$ and an <isomorphism> $\phi:P_1\otimes A_0\to P_2\otimes A_0$, set
$$
M(P_1,P_2,\phi)=\{(x_1,x_2):\phi(\bar x_1)=\bar x_2\}.
$$
For <free modules> denote the result by $M_g$. The block <matrix> identity
$$
\begin{pmatrix}g&0\\0&g^{-1}\end{pmatrix}=W(g)W(-I),\qquad W(g)=\begin{pmatrix}I&g\\0&I\end{pmatrix}\begin{pmatrix}I&0\\-g^{-1}&I\end{pmatrix}\begin{pmatrix}I&g\\0&I\end{pmatrix}
$$
expresses $\operatorname{diag}(g,g^{-1})$ as <elementary matrices>. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted <elementary matrices>, not arbitrary invertible <matrices>.
For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an <isomorphism>. The sum with $M(P_1,P_2,\phi)$ now patches two <free modules> of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an <idempotent> over a finite <free module>. Since <matrices> over the <pullback> are compatible pairs of <matrices>, applying the compatible <idempotents> to the free <pullback> reconstructs this projective from its two reductions. Compatible module <isomorphisms> patch as well. This proves <Milnor patching of projective modules> in the form needed below.
The diagonal maps in the sequence are <extension of scalars>; the two maps to $A_0$ are differences $j_{1*}-j_{2*}$. Define the connecting map by
$$
\partial[g]=[M_g]-[A^r],\qquad g\in\mathrm{GL}_r(A_0).
$$
It is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\operatorname{diag}(g,h)=\operatorname{diag}(gh,I)\operatorname{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive <homomorphism> on $K_1(A_0)$.
\b[Exactness at $K_1(A_1)\oplus K_1(A_2)$.] If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an <elementary matrix> $e$ over $A_0$. Lift $e$ to an <elementary matrix> over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible <matrix> over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.
\b[Exactness at $K_1(A_0)$.] A gluing obtained from a change of basis over either upper <ring> patches a <free module>, so the difference-map image is killed by $\partial$. Conversely, $\partial[g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual <isomorphism> $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1)^{-1}$. Therefore $[g]=j_{2*}[v_2]-j_{1*}[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.
\b[Exactness at $K_0(A)$.] Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an <isomorphism> after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial[g]$.
\b[Exactness at $K_0(A_1)\oplus K_0(A_2)$.] Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.
These arguments prove the <Mayer-Vietoris sequence in algebraic K-theory>
$$
\boxed{K_1A\to K_1A_1\oplus K_1A_2\to K_1A_0\xrightarrow{\partial}K_0A\to K_0A_1\oplus K_0A_2\to K_0A_0.}
$$
No zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.
For the <group ring> of the <cyclic group>, take the <cyclotomic Milnor square>
$$
\begin{array}{ccc}\mathbb Z[C_p]&\longrightarrow&\mathbb Z\\\downarrow&&\downarrow\\\mathbb Z[\zeta]&\longrightarrow&\mathbb F_p.\end{array}
$$
The maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta=1$, and the right map reduces modulo $p$. Since $\Phi_p(1)=p$, $\mathbb Z[\zeta]/(\zeta-1)=\mathbb F_p$. To verify the <pullback>, represent an element by a <polynomial> of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the <minimal polynomial of an algebraic element> $\Phi_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in\mathbb Z[\zeta]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]\Phi_p$ yields the pair $(a,b)$. The right map is surjective, so this is a <Milnor square>.
The hinted <cyclotomic units> are $u_k=1+\zeta+\cdots+\zeta^{k-1}$. If $kl\equiv1\pmod p$ with $1\le l<p$, the geometric sum $\sum_{j=0}^{l-1}\zeta^{kj}$ is their inverse, because $\zeta^{kl}=\zeta$. Their residues are $k\in\mathbb F_p^\times$, covering every unit of the common <field>. <Gaussian elimination> identifies $K_1(\mathbb F_p)$ with $\mathbb F_p^\times$ by <determinant>. Thus $K_1(\mathbb Z[\zeta])\to K_1(\mathbb F_p)$ is surjective, and the connecting map vanishes.
The remaining exact sequence injects $K_0(\mathbb Z[C_p])$ into $\mathbb Z\oplus K_0(\mathbb Z[\zeta])$, with image $\{(r,x):r=\operatorname{rank}x\}$. Here $K_0(\mathbb Z)=K_0(\mathbb F_p)=\mathbb Z$, and reduction of a projective over the <integral domain> $\mathbb Z[\zeta]$ has dimension its rank: localize at $(\zeta-1)$, where the projective is free, and compare its residue-field dimension with its fraction-field dimension. Projection onto the second factor is therefore a bijection on this kernel. Hence extension along the cyclotomic evaluation gives
$$
\boxed{K_0(\mathbb Z[C_p])\cong K_0(\mathbb Z[\zeta]).}
$$
The essential step is <reduction of cyclotomic units at the ramified prime>; surjectivity of a <ring> map alone would not imply surjectivity on $K_1$.
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