The center is Morita invariant, via natural endomorphisms of the identity

($\mathrm{Nat}(1,1)$ is a ring.) For natural endomorphisms $\alpha ,\beta$ of $1_{A\text{-}\mathrm{M}\mathrm{o}\mathrm{d}}$, define $\alpha +\beta$ componentwise by $(\alpha +\beta {)}_X={\alpha }_X+{\beta }_X$ in the abelian group ${\mathrm{Hom}}_A(X,X)$; this is natural because for $u:X\to Y$ both $(\alpha +\beta {)}_Y\circ u$ and $u\circ (\alpha +\beta {)}_X$ equal $u{\alpha }_X+u{\beta }_X$ by bilinearity of composition. Componentwise addition inherits associativity, commutativity, the zero transformation and additive inverses from the hom-groups, and vertical composition distributes over it on both sides because composition of $A$-linear maps is bilinear: ${\gamma }_X\circ ({\alpha }_X+{\beta }_X)={\gamma }_X{\alpha }_X+{\gamma }_X{\beta }_X$ and $({\alpha }_X+{\beta }_X)\circ {\gamma }_X={\alpha }_X{\gamma }_X+{\beta }_X{\gamma }_X$. Finally the identity $1_1$ and the zero transformation are natural. Hence $\mathrm{Nat}(1,1)$ is a ring under componentwise addition and vertical composition.

(Identification with the center.) Let $\alpha :1\Rightarrow 1$ and set $z={\alpha }_A(1)$. Left $A$-linearity gives ${\alpha }_A(a)=az$, while naturality at the left $A$-linear right multiplication $r_a:A\to A$ gives ${\alpha }_A(a)=za$. Thus $z\in Z(A)$ and ${\alpha }_A$ is a bimodule endomorphism. For $x\in X$, the left $A$-linear map ${\mathrm{\ell }}_x:A\to X$, $a\mapsto ax$, gives by naturality ${\alpha }_X(x)={\mathrm{\ell }}_x(z)=zx$. Conversely, if $z\in Z(A)$, ${\eta }_X^z(x)=zx$ is additive, satisfies $z(ax)=a(zx)$, and commutes with every $A$-linear map, so it is a natural endomorphism. The assignments $\alpha \mapsto z$ and $z\mapsto {\eta }^z$ are inverse. They preserve addition, identities, and multiplication since ${\eta }^z\circ {\eta }^w={\eta }^{zw}$. A bimodule endomorphism $f:A\to A$ similarly satisfies $f(a)=af(1)=f(1)a$, so evaluation identifies it with a unique central element and every central element supplies one. This proves both ring isomorphisms.

(Morita transport.) Let $F:A\text{-Mod}\to B\text{-Mod}$ be an additive equivalence, equipped as an adjoint equivalence with quasi-inverse $G$, unit $\eta$ and counit $\epsilon$ by [F3]. For $\alpha \in \mathrm{Nat}(1_A,1_A)$ define ${\beta }_X:={\epsilon }_X\circ F({\alpha }_{GX})\circ {\epsilon }_X^{-1}$ for $X\in B\text{-Mod}$. Each ${\beta }_X$ is an endomorphism of $X$, and $\beta$ is natural: for $u:X\to Y$, naturality of $\epsilon$ gives ${\epsilon }_Y^{-1}\circ u=FG(u)\circ {\epsilon }_X^{-1}$, hence ${\beta }_Y\circ u={\epsilon }_Y\circ F({\alpha }_{GY}\circ G(u))\circ {\epsilon }_X^{-1}={\epsilon }_Y\circ F(G(u)\circ {\alpha }_{GX})\circ {\epsilon }_X^{-1}=u\circ {\beta }_X$. The assignment $\alpha \mapsto \beta$ is additive because $F$ is additive and composition is bilinear; it carries identities to identities and composites to composites because $F$ and $\epsilon$ are functorial; and the symmetric formula ${\alpha }_Y^{\mathrm{\prime }}={\eta }_Y^{-1}\circ G({\beta }_{FY})\circ {\eta }_Y$ using $\eta$ is inverse to it, by the triangle identities and the naturality of $\eta$ and $\epsilon$. Hence it is a ring isomorphism $\mathrm{Nat}(1_A,1_A)\cong \mathrm{Nat}(1_B,1_B)$.

(Conclusion.) If $A$ and $B$ are Morita equivalent, [F4] supplies an additive equivalence $F:A\text{-Mod}\to B\text{-Mod}$, and step 1.3 gives a ring isomorphism $\mathrm{Nat}(1_A,1_A)\cong \mathrm{Nat}(1_B,1_B)$; composing with the identifications of step 1.2 gives a ring isomorphism $Z(A)\cong Z(B)$. For $A=B$ the identity functor recovers the first identification, so the statement holds in general. No element outside the given rings and functors is chosen and no choice principle is used.