Yes. Up to multiply $d$ with a scalar, we can suppose that for some $0<c\le 1$ we have $cd \le\delta\le d$ on $Y\times Y$.

Define $$d'(x,x')=\min(d(x,x'),D(x,x'));\quad \text{where}$$ $$D(x,x')=\inf_{y,y'\in Y} d(x,y)+\delta(y,y')+d(y',x').$$

Clearly $cd\le d'\le d$ on $X\times X$. It remains to prove the triangle inequality for $d'$: $d'(x,x'')\le d'(x,x')+d'(x',x'')$.

There are four cases to consider.

If $d'(x,x')=d(x,x')$ and $d'(x',x'')=d(x',x'')$, then $d'(x,x'')\le d(x,x'')\le d(x,x')+d(x',x'')=d'(x,x')+d'(x',x'')$.

Suppose $d'(x,x')=D(x,x')$ and $d'(x',x'')=d(x',x'')$. Fix $\varepsilon>0$. Choose $y,y'\in Y$ such that $d'(x,x')\ge d(x,y)+\delta(y,y')+d(y',x')-\varepsilon$. Then
$$d'(x,x')+d(x',x'')\ge d(x,y)+\delta(y,y')+d(y',x')+d(x',x'')-\varepsilon$$
$$\ge d(x,y)+\delta(y,y')+d(y',x'')-\varepsilon\ge d'(x,x'')-\varepsilon.$$
Since $\varepsilon$ is arbitrary, the inequality follows.

Case $d'(x,x')=d'(x,x')$ and $d'(x',x'')=D(x',x'')$: reduces to the previous case by switching $x$ and $x''$.

Suppose $d'(x,x')=D(x,x')$ and $d'(x',x'')=D(x',x'')$. Fix $\varepsilon>0$. Fix $y,y'_1,y'_2,y''\in Y$ such that $d'(x,x')\ge d(x,y)+\delta(y,y'_1)+d(y'_1,x')-\varepsilon$ and $d'(x',x'')\ge d(x',y'_2)+\delta(y'_2,y'')+d(y'',x'')-\varepsilon$.

So $$d'(x,x')+d'(x',x'')\ge $$
$$d(x,y)+\delta(y,y'_1)+d(y'_1,x')+d(x',y'_2)+\delta(y'_2,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d(x,y)+\delta(y,y'_1)+d(y'_1,y'_2)+\delta(y'_2,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d(x,y)+\delta(y,y'_1)+\delta(y'_1,y'_2)+\delta(y'_2,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d(x,y)+\delta(y,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d'(x,x'')-2\varepsilon.$$
Since $\varepsilon$ is arbitrary, we deduce the triangle inequality.