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Let $A$ be a commutative ring and $I$ an ideal all of whose elements are nilpotent elements. Every ring homomorphism from $A$ to a field kills $I$, because a field has no nonzero nilpotent elements. Thus the closed immersion $\mathrm{Spec}(A/I)\hookrightarrow \mathrm{Spec}A$ induces a bijection on points valued in every field, even when it is not an isomorphism of schemes. The dual numbers provide the smallest example. Testing the functor of points on all affine schemes does detect this information, by affine test schemes determine scheme morphisms.

If $X={\bigcup }_i{U}_i$ is an open cover by open subschemes, and morphisms of schemes $f_i:U_i\to Y$ agree on every overlap, there is a unique morphism of schemes $f:X\to Y$ restricting to them. The underlying continuous maps glue because their restrictions agree. The maps of structure sheaves glue using the sheaf gluing axiom, and the required local condition on stalks can be checked on the original cover. No affineness or separation assumption on $X$, $Y$, or the overlaps is needed. This is the local construction behind affine test schemes determine scheme morphisms.

Yes. First, the hypothesis implies that $f(T)$ is injective for every scheme $T$, even when $T$ is not an affine scheme. Indeed, suppose $a,b:T\to X$ satisfy $f\circ a=f\circ b$. Choose an affine open cover $T={\bigcup }_{\lambda }{V}_{\lambda }$. On each $V_{\lambda }$, injectivity of $f(V_{\lambda })$ gives $a{|}_{V_{\lambda }}=b{|}_{V_{\lambda }}$. By gluing morphisms of schemes, $a=b$.

Now choose an affine open cover $Y={\bigcup }_i{U}_i$, and denote its inclusion maps by $j_i:U_i\hookrightarrow Y$. Surjectivity of $f(U_i)$ gives morphisms of schemes $g_i:U_i\to X$ with $f\circ g_i=j_i$. On $U_i\cap U_j$, the two restrictions have the same composite with $f$. By the injectivity just established for arbitrary test schemes,In particular, the overlaps need not be affine schemes. Gluing morphisms of schemes supplies $g:Y\to X$ with $g{|}_{U_i}=g_i$, and $f\circ g=1_Y$ because this holds on every $U_i$.

Finally,Injectivity of $f(X)$, already proved without an affineness assumption, gives $g\circ f=1_X$. ThereforeThis proves the relevant instance of affine test schemes determine scheme morphisms: the restricted functor of points still retains the information needed to construct the inverse.