Prove that recursively defined function is injective and equal to subset

Actually, surjectivity is easier: We need to show that

• $f(s)=\lambda$ for some $s\in \{x,y,z\}^*$
• If $s\in f(\{x,y,z\}^*)$ then $0s\in f(\{x,y,z\}^*)$.
• If $s\in f(\{x,y,z\}^*)$ then $10s\in f(\{x,y,z\}^*)$.
• If $s\in f(\{x,y,z\}^*)$ then $11s\in f(\{x,y,z\}^*)$.

The first is clear because $f(\lambda)=\lambda$. For the other three parts, note that $s\in f(\{x,y,z\}^*)$ means $s=f(t)$ for some $t\in \{x,y,z\}^*$. But then $0s=f(xt)$, $10s=f(yt)$, $11s=f(zt)$. $\square$

For injectivity, you may need to be a bit more careful. As you suggest, we let $$ P(s)\equiv \forall t\in\{x,y,z\}^*\colon f(s)=f(t)\to s=t$$

• We have $P(\lambda)$ because if $t\ne \lambda$ then $t$ begins with $x$, $y$, or $z$, hence $f(t)$ begins with $0$, $10$, or $11$. At any rate $f(t)\ne \lambda= f(\lambda)$.
• If $s=xs'$ and $f(s)=f(t)$ then $t=\lambda$ or $t=xt'$ or $t=yt'$ or $t=zt'$. As $f(s)$ begins with $0$, we can exclude $t=\lambda$, $t=yt'$, and $t=zt'$. Hence $t=xt'$ and $0f(s')=f(s)=f(t)=0f(t')$ implies $f(s')=f(t')$. By induction hypothesis, $s'=t'$ and ultimately $s=xs'=xt'=t$.
• If $s=ys'$ and $f(s)=f(t)$ then $t=\lambda$ or $t=xt'$ or $t=yt'$ or $t=zt'$. As $f(s)$ begins with $10$, we can exclude $t=\lambda$, $t=xt'$, and $t=zt'$. Hence $t=yt'$ and $10f(s')=f(s)=f(t)=10f(t')$ implies $f(s')=f(t')$. By induction hypothesis, $s'=t'$ and ultimately $s=ys'=yt'=t$.
• If $s=zs'$ and $f(s)=f(t)$ then $t=\lambda$ or $t=xt'$ or $t=yt'$ or $t=zt'$. As $f(s)$ begins with $11$, we can exclude $t=\lambda$, $t=xt'$, and $t=yt'$. Hence $t=zt'$ and $11f(s')=f(s)=f(t)=11f(t')$ implies $f(s')=f(t')$. By induction hypothesis, $s'=t'$ and ultimately $s=zs'=zt'=t$.