Let $R$ be a ring. Find the pairs $(a,b) \in R \times R$ such that $f: R \to R$ defined by $f(x)=axb$ is a ring homomorphism.
What I want is that for $x,y \in R$ $$\begin{align*}f(x+y)&=f(x)+f(y) \\ a(x+y)b &= axb+ayb \\ (ax+ay)b &= axb+ayb \\ axb+ayb &= axb +ayb\end{align*}$$ so I think this condition is satisfied for all $(a,b ) \in R \times R$. The other condition is that $$\begin{align*}f(xy)&=f(x)f(y) \\ axyb &= axbayb\end{align*}$$ and I think that this is only satisfied whenever $ba = e$ as then $axbayb = axyb$. So $a$ must be the inverse of $b$ or the other way around. The pairs should be then $(a,a^{-1}), (b,b^{-1})$?
In a ring with unity you also want the morphism to respect the unity. The condition $b=a^{-1}$ satisfies this.
From the unity condition $f(e)=e$ for the multiplicative unity $e$ it is easy to see, that $ba=e$ is necessary. In case $b$ (and thus $a$) are invertible this implies $b=a^{-1}$. In general, as Adayah pointed out, not all left inverses need to be right inverses, i.e. they need not be invertible.
From the example in the link ($R=\{g:\mathbb{Z}\rightarrow\mathbb{Z}\}$ selfmaps of the integers), where we have elements with left inverses that do not have right inverses (e.g. the doubling map described on that page), you can construct morphisms $f$ where $b\neq a^{-1}$, because $a^{-1}$ does not exist.
In rings without unity, this is no longer the case. Even the inverse does not make sense there. As an example you can look at the $2$-dimensional $k$-algebra $S=ku\oplus kv$ given with multiplication $u^2=v, uv=vu=v^2=0$. You get the $0$-morphism for any choice of $a,b$, but it is a homomorphism of rings without unit.
Here's a less trivial example. The $3$-dimensional $k$-algebra $S=ku\oplus kv\oplus kw$ given with multiplication $u^2=v,$ $uv=vu=w,$ $uw=v^2=vw=wu=wv=w^2=0$. You get a morphism with image $kw$ for the choice of $a=b=u$.