Let $T = \{T_n; n=0,1,\ldots\}$ be a renewal sequence with inter-renewal distribuition $F.$
Let $$A = \{A(t); t\geq 0\}$$ and $$B = \{B(t); t\geq 0\}$$ be the forward and backward recurrences time respectively. Find a renewal equation for $$h(t) = P\{A(t) >x, B(t) >y\}.$$
My Take: I am able to figure out that the renewal equation for $A$ is obtained as $$P\{A(t) > x, T_1 \leq t \} = \int_0^t P\{A(t-u)>x\} dF(u)$$ such that the renewal equation is $$h(t)_A = 1 - F(t+x) + F \star h(t).$$
Similarly for $B$ it is obtained as follows, $$P\{B(t) > x, T_1 \leq t \} = \int_0^t P\{B(t-u)>x\} dF(u)$$ and $$h(t)_B = (t>x) (1-F(t))+ F \star h(t).$$
Thus, I am not sure whether combining $h(t)_A$ and $h(t)_B$ is legit for the joint renewal equation for $A$ and $B$ i.e., $h(t) = P\{A(t) >x, B(t) >y\}.$ I welcome any ideas to this problem. You could provide your solution as well.
Conditioning on the time of the first renewal is the right way to start, but your results don't seem correct. For $s\geqslant 0$ consider $$ \mathbb P(A(t)>y) = \int_0^\infty \mathbb P(A(t)>y\mid T_1=s)\ \mathsf d F(s). $$ For $s\in[0,t]$ the integrand above is equal to $\mathbb P(A(t-s)>y)$, for $s\in (t,t+y]$ it is zero, and for $s>t+y$ it is one. Hence $$ \mathbb P(A(t)>y) = \int_0^t \mathbb P(A(t-s)>y)\ \mathsf d F(s)+\int_{t+y}^\infty \ \mathsf d F(s) = 1-F(t+y)+h_A(y)\star F,$$ where the renewal equation is $h_A(y) = 1-F(t+y)+h_A(y)$.
For the backward recurrence time, note that $\mathbb P(B(t)\geqslant x) = \mathbb P(A(t-x)>x)$ for $x\in[0,t]$, from which it follows that $$ \mathbb P(B(t)\geqslant x) = 1-F(t+\int_0^{t-x}(1-F(t-s))\ \mathsf d m(s). $$ Finally, $\mathbb P(B(t)\geqslant x,A(t)>y) = \mathbb P(A_{t-x}>x+y)$. From the above it follows that $$ \mathbb P(B(t)\geqslant x, A(t)>y)=1-F(t+y)+\int_0^{t-x}(1-(F(t+y-s))\ \mathsf d m(s),\quad 0\leqslant x\leqslant t, y\geqslant 0. $$